Multiply two logs
$begingroup$
How can we multiply two logs or how can we simplify them for example:
$$ln(x)*ln(1-x)$$
calculus logarithms
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
$endgroup$
add a comment |
$begingroup$
How can we multiply two logs or how can we simplify them for example:
$$ln(x)*ln(1-x)$$
calculus logarithms
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
$endgroup$
add a comment |
$begingroup$
How can we multiply two logs or how can we simplify them for example:
$$ln(x)*ln(1-x)$$
calculus logarithms
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
$endgroup$
How can we multiply two logs or how can we simplify them for example:
$$ln(x)*ln(1-x)$$
calculus logarithms
calculus logarithms
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
asked Jan 12 at 11:43
Weez KhanWeez Khan
135
135
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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You can not combine these terms.
Only in the case $$lnleft(frac{a}{b}right)=ln(a)-ln(b)$$ and $$ln(ab)=ln(a)+ln(b)$$ if $$a,b>0$$
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
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add a comment |
$begingroup$
They're already in the simplest form. So the expression can't be simplified.
But you could combine them if you didn't mind it getting more complicated. I don't recommend it unless you've got a special use for the result:
$$ln(a)cdotln(b)=ln(a^{ln(b)})=ln(b^{ln(a)})$$
So it's not been simplified, just complicated in two different ways. But for the fun of it, let's equate the two ways:
$$ln(a^{ln(b)})=ln(b^{ln(a)})$$
then exponentiate both sides to get a nice identity:
$$a^{ln(b)}=b^{ln(a)}$$
which isn't what you were trying to do, but is a nice side effect of playing with it.
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
$ln(x)+ln(1-x) = ln(1-x^2)$ but
$$ln(x)*ln(1-x)$$ cannot be multiplied or simplified further other way round anymore.
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
You can not combine these terms.
Only in the case $$lnleft(frac{a}{b}right)=ln(a)-ln(b)$$ and $$ln(ab)=ln(a)+ln(b)$$ if $$a,b>0$$
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
You can not combine these terms.
Only in the case $$lnleft(frac{a}{b}right)=ln(a)-ln(b)$$ and $$ln(ab)=ln(a)+ln(b)$$ if $$a,b>0$$
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
add a comment |
$begingroup$
You can not combine these terms.
Only in the case $$lnleft(frac{a}{b}right)=ln(a)-ln(b)$$ and $$ln(ab)=ln(a)+ln(b)$$ if $$a,b>0$$
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
You can not combine these terms.
Only in the case $$lnleft(frac{a}{b}right)=ln(a)-ln(b)$$ and $$ln(ab)=ln(a)+ln(b)$$ if $$a,b>0$$
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 12 at 11:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
add a comment |
add a comment |
$begingroup$
They're already in the simplest form. So the expression can't be simplified.
But you could combine them if you didn't mind it getting more complicated. I don't recommend it unless you've got a special use for the result:
$$ln(a)cdotln(b)=ln(a^{ln(b)})=ln(b^{ln(a)})$$
So it's not been simplified, just complicated in two different ways. But for the fun of it, let's equate the two ways:
$$ln(a^{ln(b)})=ln(b^{ln(a)})$$
then exponentiate both sides to get a nice identity:
$$a^{ln(b)}=b^{ln(a)}$$
which isn't what you were trying to do, but is a nice side effect of playing with it.
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
They're already in the simplest form. So the expression can't be simplified.
But you could combine them if you didn't mind it getting more complicated. I don't recommend it unless you've got a special use for the result:
$$ln(a)cdotln(b)=ln(a^{ln(b)})=ln(b^{ln(a)})$$
So it's not been simplified, just complicated in two different ways. But for the fun of it, let's equate the two ways:
$$ln(a^{ln(b)})=ln(b^{ln(a)})$$
then exponentiate both sides to get a nice identity:
$$a^{ln(b)}=b^{ln(a)}$$
which isn't what you were trying to do, but is a nice side effect of playing with it.
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
$endgroup$
add a comment |
$begingroup$
They're already in the simplest form. So the expression can't be simplified.
But you could combine them if you didn't mind it getting more complicated. I don't recommend it unless you've got a special use for the result:
$$ln(a)cdotln(b)=ln(a^{ln(b)})=ln(b^{ln(a)})$$
So it's not been simplified, just complicated in two different ways. But for the fun of it, let's equate the two ways:
$$ln(a^{ln(b)})=ln(b^{ln(a)})$$
then exponentiate both sides to get a nice identity:
$$a^{ln(b)}=b^{ln(a)}$$
which isn't what you were trying to do, but is a nice side effect of playing with it.
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
$endgroup$
They're already in the simplest form. So the expression can't be simplified.
But you could combine them if you didn't mind it getting more complicated. I don't recommend it unless you've got a special use for the result:
$$ln(a)cdotln(b)=ln(a^{ln(b)})=ln(b^{ln(a)})$$
So it's not been simplified, just complicated in two different ways. But for the fun of it, let's equate the two ways:
$$ln(a^{ln(b)})=ln(b^{ln(a)})$$
then exponentiate both sides to get a nice identity:
$$a^{ln(b)}=b^{ln(a)}$$
which isn't what you were trying to do, but is a nice side effect of playing with it.
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
answered Jan 12 at 15:35
timtfjtimtfj
2,468420
2,468420
add a comment |
add a comment |
$begingroup$
$ln(x)+ln(1-x) = ln(1-x^2)$ but
$$ln(x)*ln(1-x)$$ cannot be multiplied or simplified further other way round anymore.
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
$ln(x)+ln(1-x) = ln(1-x^2)$ but
$$ln(x)*ln(1-x)$$ cannot be multiplied or simplified further other way round anymore.
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
$endgroup$
add a comment |
$begingroup$
$ln(x)+ln(1-x) = ln(1-x^2)$ but
$$ln(x)*ln(1-x)$$ cannot be multiplied or simplified further other way round anymore.
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
$endgroup$
$ln(x)+ln(1-x) = ln(1-x^2)$ but
$$ln(x)*ln(1-x)$$ cannot be multiplied or simplified further other way round anymore.
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
answered Jan 12 at 15:40
NarasimhamNarasimham
20.9k62158
20.9k62158
add a comment |
add a comment |
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