How to prove this equation $log^2_a(x) = log_a(x^{log_ax})$?
$begingroup$
I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?
functions logarithms
edited Dec 29 '18 at 16:21
Bernard
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
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add a comment |
$begingroup$
I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?
functions logarithms
edited Dec 29 '18 at 16:21
Bernard
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
$endgroup$
add a comment |
$begingroup$
I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?
functions logarithms
edited Dec 29 '18 at 16:21
Bernard
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
$endgroup$
I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?
functions logarithms
functions logarithms
edited Dec 29 '18 at 16:21
Bernard
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
edited Dec 29 '18 at 16:21
Bernard
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
edited Dec 29 '18 at 16:21
Bernard
119k639112
edited Dec 29 '18 at 16:21
Bernard
119k639112
edited Dec 29 '18 at 16:21
Bernard
119k639112
119k639112
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
asked Dec 29 '18 at 15:58
Vitalii PaprotskyiVitalii Paprotskyi
6517
6517
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
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$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
add a comment |
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Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
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add a comment |
$begingroup$
Recall the following property of logarithms:
$$log_a b^c = clog_a b; quad b > 0$$
Applying it to the right-hand side, you get
$$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
$endgroup$
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
add a comment |
$begingroup$
Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
$endgroup$
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
add a comment |
$begingroup$
Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
$endgroup$
Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
answered Dec 29 '18 at 16:00
WuestenfuxWuestenfux
3,8311411
3,8311411
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
add a comment |
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
$begingroup$
oh, of course, I completely forgot about it... Thanks!
$endgroup$
– Vitalii Paprotskyi
Dec 29 '18 at 16:03
add a comment |
$begingroup$
Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
add a comment |
$begingroup$
Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
add a comment |
$begingroup$
Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
73.5k42865
add a comment |
add a comment |
$begingroup$
Recall the following property of logarithms:
$$log_a b^c = clog_a b; quad b > 0$$
Applying it to the right-hand side, you get
$$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
$endgroup$
add a comment |
$begingroup$
Recall the following property of logarithms:
$$log_a b^c = clog_a b; quad b > 0$$
Applying it to the right-hand side, you get
$$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
$endgroup$
add a comment |
$begingroup$
Recall the following property of logarithms:
$$log_a b^c = clog_a b; quad b > 0$$
Applying it to the right-hand side, you get
$$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
$endgroup$
Recall the following property of logarithms:
$$log_a b^c = clog_a b; quad b > 0$$
Applying it to the right-hand side, you get
$$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
answered Dec 29 '18 at 16:01
KM101KM101
5,8711423
5,8711423
add a comment |
add a comment |
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