Easy way to compute logarithms without a calculator?
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I would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.
$$log_8128 = frac 73$$
How do you do this?
logarithms
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
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|
show 1 more comment
$begingroup$
I would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.
$$log_8128 = frac 73$$
How do you do this?
logarithms
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
$endgroup$
5
$begingroup$
If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
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– Ross Millikan
Feb 13 '16 at 16:13
2
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You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
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– Von Neumann
Feb 13 '16 at 23:16
1
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@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
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– zahbaz
Feb 15 '16 at 1:19
2
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slide rule works for me...
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– Joel
Feb 15 '16 at 10:03
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@Joel he said without a calculator :-)
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– Matt Gutting
Feb 15 '16 at 12:17
|
show 1 more comment
$begingroup$
I would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.
$$log_8128 = frac 73$$
How do you do this?
logarithms
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
$endgroup$
I would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.
$$log_8128 = frac 73$$
How do you do this?
logarithms
logarithms
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
edited Feb 15 '16 at 13:41
Najib Idrissi
40.9k471139
40.9k471139
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
asked Feb 13 '16 at 16:10
Balázs VinczeBalázs Vincze
289137
289137
5
$begingroup$
If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
$endgroup$
– Ross Millikan
Feb 13 '16 at 16:13
2
$begingroup$
You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
$endgroup$
– Von Neumann
Feb 13 '16 at 23:16
1
$begingroup$
@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
$endgroup$
– zahbaz
Feb 15 '16 at 1:19
2
$begingroup$
slide rule works for me...
$endgroup$
– Joel
Feb 15 '16 at 10:03
$begingroup$
@Joel he said without a calculator :-)
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:17
|
show 1 more comment
5
$begingroup$
If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
$endgroup$
– Ross Millikan
Feb 13 '16 at 16:13
2
$begingroup$
You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
$endgroup$
– Von Neumann
Feb 13 '16 at 23:16
1
$begingroup$
@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
$endgroup$
– zahbaz
Feb 15 '16 at 1:19
2
$begingroup$
slide rule works for me...
$endgroup$
– Joel
Feb 15 '16 at 10:03
$begingroup$
@Joel he said without a calculator :-)
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:17
5
5
$begingroup$
If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
$endgroup$
– Ross Millikan
Feb 13 '16 at 16:13
$begingroup$
If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
$endgroup$
– Ross Millikan
Feb 13 '16 at 16:13
2
2
$begingroup$
You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
$endgroup$
– Von Neumann
Feb 13 '16 at 23:16
$begingroup$
You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
$endgroup$
– Von Neumann
Feb 13 '16 at 23:16
1
1
$begingroup$
@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
$endgroup$
– zahbaz
Feb 15 '16 at 1:19
$begingroup$
@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
$endgroup$
– zahbaz
Feb 15 '16 at 1:19
2
2
$begingroup$
slide rule works for me...
$endgroup$
– Joel
Feb 15 '16 at 10:03
$begingroup$
slide rule works for me...
$endgroup$
– Joel
Feb 15 '16 at 10:03
$begingroup$
@Joel he said without a calculator :-)
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:17
$begingroup$
@Joel he said without a calculator :-)
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:17
|
show 1 more comment
10 Answers
10
active
oldest
votes
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To evaluate $log_8 128$, let
$$log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
begin{align*}
(2^3)^x & = 2^7\
2^{3x} & = 2^7
end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
begin{align*}
3x & = 7\
x & = frac{7}{3}
end{align*}
Check: If $x = frac{7}{3}$, then
$$8^x = 8^{frac{7}{3}} = (8^{frac{1}{3}})^7 = 2^7 = 128$$
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
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add a comment |
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Using $log_xy=dfrac{log_ay}{log_ax}$ and $log(z^m)=mlog z$ where all the logarithms must remain defined unlike $log_a1nelog_a(-1)^2$
$$log_8{128}=dfrac{log_a(2^7)}{log_a(2^3)}=dfrac{7log_a2}{3log_a2}=?$$
Clearly, $log_a2$ is non-zero finite for finite real $a>0,ne1$
See Laws of Logarithms
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
$endgroup$
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
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– richard1941
Mar 6 '18 at 17:16
add a comment |
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Another way of doing this:
$$ 128= 2^7 = (2^3)^frac{7}{3} = 8^frac{7}{3}$$
$$ log_8 128 = log_8 (8)^frac{7}{3} = frac{7}{3}$$
Note the laws of logarithm used here: $$ log_a a = 1$$
$$ log_y x^a= a log_y x$$
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
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add a comment |
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In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $frac{c}a$.
For example,
$$log_{8}(128) = log_{2^3}(2^7)=log_{2^3}((2^3)^{frac73})=frac73$$
$$log_{27}(2187) = log_{3^3}(3^7)=log_{3^3}((3^3)^{frac73})=frac73$$
$$log_{36}(216) = log_{6^2}(6^3)=log_{6^2}((6^2)^{frac32})=frac32$$
edited Sep 26 '17 at 12:34
Mr Reality
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
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All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $sqrt 3 $ note that:
$ 7^2 = 49 approx 48 = 3 * 16 = 3 * 4 ^ 2$
and so
$sqrt 3 approx 7 / 4 = 1.75 $
Similarly for $sqrt 2 $ note that $ 10 ^ 2 = 100 approx 2 * 49 = 2 * 7 ^ 2 $ and so $sqrt 2 approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ sqrt 2 / 1.4 approx 1.42857 $ and so a better approximation is $ sqrt 2 approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ sqrt 2 approx 1.41421356 $, which is as accurate as many hand calculators.
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
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add a comment |
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This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
Definition of logarithm in reals may help: $log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without calculator. I will tell you a method that I use: since $exp 3 approx 20$, you can take $log 20 approx 3$. Hence, to calculate $log n$ in practical applications, first calculate $log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\ log 25600000000 approx 8 cdot 3 = 24\ log 34627486221 = log 25600000000 + log (34627486221 / 25600000000) approx 24 + log 1.35 approx 24.35$$ in which the relative error is less than $1/297$.
Hope this helps.
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add a comment |
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Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). If n is a power of 2, you get to take a lot of square roots. I gave a paper last year that discusses ways to improve on this idea and offers improved interpolation in log tables, in case civilization is blasted back to the pre-calculator days by GMO, gluten, oxidants, and inorganic farming. Available upon request.
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
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Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
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– richard1941
Mar 6 '18 at 17:23
add a comment |
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In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of log(2) is given, obtaining the result
0.6921 < log(2) < 0.6935 “with very little effort”, as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!”
-- Carl Friedrich Gauss
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
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add a comment |
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Use the logarithm rules to simplify
$$log frac{5}{36}−logfrac{5}{9}$$
Choose the best available answer below
$$-log 9$$
$$phantom{-}log 5$$
$$phantom{-}log 4$$
$$-log 4$$
edited Dec 30 '18 at 3:24
dantopa
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
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add a comment |
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10 Answers
10
active
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10 Answers
10
active
oldest
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active
oldest
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active
oldest
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$begingroup$
To evaluate $log_8 128$, let
$$log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
begin{align*}
(2^3)^x & = 2^7\
2^{3x} & = 2^7
end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
begin{align*}
3x & = 7\
x & = frac{7}{3}
end{align*}
Check: If $x = frac{7}{3}$, then
$$8^x = 8^{frac{7}{3}} = (8^{frac{1}{3}})^7 = 2^7 = 128$$
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
$endgroup$
add a comment |
$begingroup$
To evaluate $log_8 128$, let
$$log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
begin{align*}
(2^3)^x & = 2^7\
2^{3x} & = 2^7
end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
begin{align*}
3x & = 7\
x & = frac{7}{3}
end{align*}
Check: If $x = frac{7}{3}$, then
$$8^x = 8^{frac{7}{3}} = (8^{frac{1}{3}})^7 = 2^7 = 128$$
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
$endgroup$
add a comment |
$begingroup$
To evaluate $log_8 128$, let
$$log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
begin{align*}
(2^3)^x & = 2^7\
2^{3x} & = 2^7
end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
begin{align*}
3x & = 7\
x & = frac{7}{3}
end{align*}
Check: If $x = frac{7}{3}$, then
$$8^x = 8^{frac{7}{3}} = (8^{frac{1}{3}})^7 = 2^7 = 128$$
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
$endgroup$
To evaluate $log_8 128$, let
$$log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
begin{align*}
(2^3)^x & = 2^7\
2^{3x} & = 2^7
end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
begin{align*}
3x & = 7\
x & = frac{7}{3}
end{align*}
Check: If $x = frac{7}{3}$, then
$$8^x = 8^{frac{7}{3}} = (8^{frac{1}{3}})^7 = 2^7 = 128$$
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
edited Feb 13 '16 at 23:09
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
answered Feb 13 '16 at 16:19
N. F. TaussigN. F. Taussig
43.8k93355
43.8k93355
add a comment |
add a comment |
$begingroup$
Using $log_xy=dfrac{log_ay}{log_ax}$ and $log(z^m)=mlog z$ where all the logarithms must remain defined unlike $log_a1nelog_a(-1)^2$
$$log_8{128}=dfrac{log_a(2^7)}{log_a(2^3)}=dfrac{7log_a2}{3log_a2}=?$$
Clearly, $log_a2$ is non-zero finite for finite real $a>0,ne1$
See Laws of Logarithms
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Using $log_xy=dfrac{log_ay}{log_ax}$ and $log(z^m)=mlog z$ where all the logarithms must remain defined unlike $log_a1nelog_a(-1)^2$
$$log_8{128}=dfrac{log_a(2^7)}{log_a(2^3)}=dfrac{7log_a2}{3log_a2}=?$$
Clearly, $log_a2$ is non-zero finite for finite real $a>0,ne1$
See Laws of Logarithms
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
Using $log_xy=dfrac{log_ay}{log_ax}$ and $log(z^m)=mlog z$ where all the logarithms must remain defined unlike $log_a1nelog_a(-1)^2$
$$log_8{128}=dfrac{log_a(2^7)}{log_a(2^3)}=dfrac{7log_a2}{3log_a2}=?$$
Clearly, $log_a2$ is non-zero finite for finite real $a>0,ne1$
See Laws of Logarithms
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Using $log_xy=dfrac{log_ay}{log_ax}$ and $log(z^m)=mlog z$ where all the logarithms must remain defined unlike $log_a1nelog_a(-1)^2$
$$log_8{128}=dfrac{log_a(2^7)}{log_a(2^3)}=dfrac{7log_a2}{3log_a2}=?$$
Clearly, $log_a2$ is non-zero finite for finite real $a>0,ne1$
See Laws of Logarithms
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
edited Feb 14 '16 at 3:23
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
answered Feb 13 '16 at 16:13
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
$begingroup$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
$endgroup$
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
add a comment |
$begingroup$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
$endgroup$
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
add a comment |
$begingroup$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
$endgroup$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
answered Feb 14 '16 at 2:45
personjerrypersonjerry
23113
23113
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
add a comment |
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
7
7
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
The slide rule is a calculator. A mechanical calculator.
$endgroup$
– Matt Gutting
Feb 15 '16 at 12:18
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
$begingroup$
So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table.
$endgroup$
– richard1941
Mar 6 '18 at 17:16
add a comment |
$begingroup$
Another way of doing this:
$$ 128= 2^7 = (2^3)^frac{7}{3} = 8^frac{7}{3}$$
$$ log_8 128 = log_8 (8)^frac{7}{3} = frac{7}{3}$$
Note the laws of logarithm used here: $$ log_a a = 1$$
$$ log_y x^a= a log_y x$$
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
$endgroup$
add a comment |
$begingroup$
Another way of doing this:
$$ 128= 2^7 = (2^3)^frac{7}{3} = 8^frac{7}{3}$$
$$ log_8 128 = log_8 (8)^frac{7}{3} = frac{7}{3}$$
Note the laws of logarithm used here: $$ log_a a = 1$$
$$ log_y x^a= a log_y x$$
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
$endgroup$
add a comment |
$begingroup$
Another way of doing this:
$$ 128= 2^7 = (2^3)^frac{7}{3} = 8^frac{7}{3}$$
$$ log_8 128 = log_8 (8)^frac{7}{3} = frac{7}{3}$$
Note the laws of logarithm used here: $$ log_a a = 1$$
$$ log_y x^a= a log_y x$$
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
$endgroup$
Another way of doing this:
$$ 128= 2^7 = (2^3)^frac{7}{3} = 8^frac{7}{3}$$
$$ log_8 128 = log_8 (8)^frac{7}{3} = frac{7}{3}$$
Note the laws of logarithm used here: $$ log_a a = 1$$
$$ log_y x^a= a log_y x$$
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
answered Feb 13 '16 at 16:22
John_dydxJohn_dydx
3,7111924
3,7111924
add a comment |
add a comment |
$begingroup$
In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $frac{c}a$.
For example,
$$log_{8}(128) = log_{2^3}(2^7)=log_{2^3}((2^3)^{frac73})=frac73$$
$$log_{27}(2187) = log_{3^3}(3^7)=log_{3^3}((3^3)^{frac73})=frac73$$
$$log_{36}(216) = log_{6^2}(6^3)=log_{6^2}((6^2)^{frac32})=frac32$$
edited Sep 26 '17 at 12:34
Mr Reality
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
$endgroup$
add a comment |
$begingroup$
In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $frac{c}a$.
For example,
$$log_{8}(128) = log_{2^3}(2^7)=log_{2^3}((2^3)^{frac73})=frac73$$
$$log_{27}(2187) = log_{3^3}(3^7)=log_{3^3}((3^3)^{frac73})=frac73$$
$$log_{36}(216) = log_{6^2}(6^3)=log_{6^2}((6^2)^{frac32})=frac32$$
edited Sep 26 '17 at 12:34
Mr Reality
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
$endgroup$
add a comment |
$begingroup$
In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $frac{c}a$.
For example,
$$log_{8}(128) = log_{2^3}(2^7)=log_{2^3}((2^3)^{frac73})=frac73$$
$$log_{27}(2187) = log_{3^3}(3^7)=log_{3^3}((3^3)^{frac73})=frac73$$
$$log_{36}(216) = log_{6^2}(6^3)=log_{6^2}((6^2)^{frac32})=frac32$$
edited Sep 26 '17 at 12:34
Mr Reality
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
$endgroup$
In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $frac{c}a$.
For example,
$$log_{8}(128) = log_{2^3}(2^7)=log_{2^3}((2^3)^{frac73})=frac73$$
$$log_{27}(2187) = log_{3^3}(3^7)=log_{3^3}((3^3)^{frac73})=frac73$$
$$log_{36}(216) = log_{6^2}(6^3)=log_{6^2}((6^2)^{frac32})=frac32$$
edited Sep 26 '17 at 12:34
Mr Reality
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
edited Sep 26 '17 at 12:34
Mr Reality
14610
edited Sep 26 '17 at 12:34
Mr Reality
14610
edited Sep 26 '17 at 12:34
Mr Reality
14610
14610
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
answered Feb 13 '16 at 16:28
wythagoraswythagoras
21.6k444104
21.6k444104
add a comment |
add a comment |
$begingroup$
All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $sqrt 3 $ note that:
$ 7^2 = 49 approx 48 = 3 * 16 = 3 * 4 ^ 2$
and so
$sqrt 3 approx 7 / 4 = 1.75 $
Similarly for $sqrt 2 $ note that $ 10 ^ 2 = 100 approx 2 * 49 = 2 * 7 ^ 2 $ and so $sqrt 2 approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ sqrt 2 / 1.4 approx 1.42857 $ and so a better approximation is $ sqrt 2 approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ sqrt 2 approx 1.41421356 $, which is as accurate as many hand calculators.
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
$endgroup$
add a comment |
$begingroup$
All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $sqrt 3 $ note that:
$ 7^2 = 49 approx 48 = 3 * 16 = 3 * 4 ^ 2$
and so
$sqrt 3 approx 7 / 4 = 1.75 $
Similarly for $sqrt 2 $ note that $ 10 ^ 2 = 100 approx 2 * 49 = 2 * 7 ^ 2 $ and so $sqrt 2 approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ sqrt 2 / 1.4 approx 1.42857 $ and so a better approximation is $ sqrt 2 approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ sqrt 2 approx 1.41421356 $, which is as accurate as many hand calculators.
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
$endgroup$
add a comment |
$begingroup$
All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $sqrt 3 $ note that:
$ 7^2 = 49 approx 48 = 3 * 16 = 3 * 4 ^ 2$
and so
$sqrt 3 approx 7 / 4 = 1.75 $
Similarly for $sqrt 2 $ note that $ 10 ^ 2 = 100 approx 2 * 49 = 2 * 7 ^ 2 $ and so $sqrt 2 approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ sqrt 2 / 1.4 approx 1.42857 $ and so a better approximation is $ sqrt 2 approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ sqrt 2 approx 1.41421356 $, which is as accurate as many hand calculators.
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
$endgroup$
All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $sqrt 3 $ note that:
$ 7^2 = 49 approx 48 = 3 * 16 = 3 * 4 ^ 2$
and so
$sqrt 3 approx 7 / 4 = 1.75 $
Similarly for $sqrt 2 $ note that $ 10 ^ 2 = 100 approx 2 * 49 = 2 * 7 ^ 2 $ and so $sqrt 2 approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ sqrt 2 / 1.4 approx 1.42857 $ and so a better approximation is $ sqrt 2 approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ sqrt 2 approx 1.41421356 $, which is as accurate as many hand calculators.
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
edited Feb 15 '16 at 8:48
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
answered Feb 15 '16 at 8:33
Pieter GeerkensPieter Geerkens
792410
792410
add a comment |
add a comment |
$begingroup$
This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
Definition of logarithm in reals may help: $log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without calculator. I will tell you a method that I use: since $exp 3 approx 20$, you can take $log 20 approx 3$. Hence, to calculate $log n$ in practical applications, first calculate $log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\ log 25600000000 approx 8 cdot 3 = 24\ log 34627486221 = log 25600000000 + log (34627486221 / 25600000000) approx 24 + log 1.35 approx 24.35$$ in which the relative error is less than $1/297$.
Hope this helps.
$endgroup$
add a comment |
$begingroup$
This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
Definition of logarithm in reals may help: $log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without calculator. I will tell you a method that I use: since $exp 3 approx 20$, you can take $log 20 approx 3$. Hence, to calculate $log n$ in practical applications, first calculate $log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\ log 25600000000 approx 8 cdot 3 = 24\ log 34627486221 = log 25600000000 + log (34627486221 / 25600000000) approx 24 + log 1.35 approx 24.35$$ in which the relative error is less than $1/297$.
Hope this helps.
$endgroup$
add a comment |
$begingroup$
This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
Definition of logarithm in reals may help: $log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without calculator. I will tell you a method that I use: since $exp 3 approx 20$, you can take $log 20 approx 3$. Hence, to calculate $log n$ in practical applications, first calculate $log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\ log 25600000000 approx 8 cdot 3 = 24\ log 34627486221 = log 25600000000 + log (34627486221 / 25600000000) approx 24 + log 1.35 approx 24.35$$ in which the relative error is less than $1/297$.
Hope this helps.
$endgroup$
This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
Definition of logarithm in reals may help: $log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without calculator. I will tell you a method that I use: since $exp 3 approx 20$, you can take $log 20 approx 3$. Hence, to calculate $log n$ in practical applications, first calculate $log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\ log 25600000000 approx 8 cdot 3 = 24\ log 34627486221 = log 25600000000 + log (34627486221 / 25600000000) approx 24 + log 1.35 approx 24.35$$ in which the relative error is less than $1/297$.
Hope this helps.
edited Feb 15 '16 at 14:49
answered Feb 14 '16 at 14:01
user98186
add a comment |
add a comment |
$begingroup$
Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). If n is a power of 2, you get to take a lot of square roots. I gave a paper last year that discusses ways to improve on this idea and offers improved interpolation in log tables, in case civilization is blasted back to the pre-calculator days by GMO, gluten, oxidants, and inorganic farming. Available upon request.
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
$endgroup$
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
add a comment |
$begingroup$
Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). If n is a power of 2, you get to take a lot of square roots. I gave a paper last year that discusses ways to improve on this idea and offers improved interpolation in log tables, in case civilization is blasted back to the pre-calculator days by GMO, gluten, oxidants, and inorganic farming. Available upon request.
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
$endgroup$
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
add a comment |
$begingroup$
Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). If n is a power of 2, you get to take a lot of square roots. I gave a paper last year that discusses ways to improve on this idea and offers improved interpolation in log tables, in case civilization is blasted back to the pre-calculator days by GMO, gluten, oxidants, and inorganic farming. Available upon request.
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
$endgroup$
Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). If n is a power of 2, you get to take a lot of square roots. I gave a paper last year that discusses ways to improve on this idea and offers improved interpolation in log tables, in case civilization is blasted back to the pre-calculator days by GMO, gluten, oxidants, and inorganic farming. Available upon request.
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
answered Mar 6 '18 at 17:07
richard1941richard1941
50629
50629
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
add a comment |
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
$begingroup$
Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong.
$endgroup$
– richard1941
Mar 6 '18 at 17:23
add a comment |
$begingroup$
In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of log(2) is given, obtaining the result
0.6921 < log(2) < 0.6935 “with very little effort”, as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!”
-- Carl Friedrich Gauss
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
$endgroup$
add a comment |
$begingroup$
In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of log(2) is given, obtaining the result
0.6921 < log(2) < 0.6935 “with very little effort”, as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!”
-- Carl Friedrich Gauss
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
$endgroup$
add a comment |
$begingroup$
In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of log(2) is given, obtaining the result
0.6921 < log(2) < 0.6935 “with very little effort”, as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!”
-- Carl Friedrich Gauss
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
$endgroup$
In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of log(2) is given, obtaining the result
0.6921 < log(2) < 0.6935 “with very little effort”, as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!”
-- Carl Friedrich Gauss
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
answered Feb 17 '16 at 15:42
Mike JonesMike Jones
2,50412735
2,50412735
add a comment |
add a comment |
$begingroup$
Use the logarithm rules to simplify
$$log frac{5}{36}−logfrac{5}{9}$$
Choose the best available answer below
$$-log 9$$
$$phantom{-}log 5$$
$$phantom{-}log 4$$
$$-log 4$$
edited Dec 30 '18 at 3:24
dantopa
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
$endgroup$
add a comment |
$begingroup$
Use the logarithm rules to simplify
$$log frac{5}{36}−logfrac{5}{9}$$
Choose the best available answer below
$$-log 9$$
$$phantom{-}log 5$$
$$phantom{-}log 4$$
$$-log 4$$
edited Dec 30 '18 at 3:24
dantopa
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
$endgroup$
add a comment |
$begingroup$
Use the logarithm rules to simplify
$$log frac{5}{36}−logfrac{5}{9}$$
Choose the best available answer below
$$-log 9$$
$$phantom{-}log 5$$
$$phantom{-}log 4$$
$$-log 4$$
edited Dec 30 '18 at 3:24
dantopa
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
$endgroup$
Use the logarithm rules to simplify
$$log frac{5}{36}−logfrac{5}{9}$$
Choose the best available answer below
$$-log 9$$
$$phantom{-}log 5$$
$$phantom{-}log 4$$
$$-log 4$$
edited Dec 30 '18 at 3:24
dantopa
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
edited Dec 30 '18 at 3:24
dantopa
6,44932142
edited Dec 30 '18 at 3:24
dantopa
6,44932142
edited Dec 30 '18 at 3:24
dantopa
6,44932142
6,44932142
answered Dec 29 '18 at 23:38
AvrilAvril
11
answered Dec 29 '18 at 23:38
AvrilAvril
11
answered Dec 29 '18 at 23:38
AvrilAvril
11
11
add a comment |
add a comment |
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If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these.
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– Ross Millikan
Feb 13 '16 at 16:13
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You could have written a more interesting example. What about $ln(200.34)$ or $log_{11}(4)$?
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– Von Neumann
Feb 13 '16 at 23:16
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@KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm.
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– zahbaz
Feb 15 '16 at 1:19
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slide rule works for me...
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– Joel
Feb 15 '16 at 10:03
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@Joel he said without a calculator :-)
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– Matt Gutting
Feb 15 '16 at 12:17