Logarithm proof problem: $a^{\log_b c} = c^{\log_b a}$

Question

I have been hit with a homework problem that I just have no idea how to approach. Any help from you all is very much appreciated. Here is the problem

Prove the equation: $a^{\log_b c} = c^{\log_b a}$

Any ideas?

Answer 1

$\large{a^{\log_b c}=e^{\ln a \cdot\log_b c}= e^{\ln a\cdot\ln c/\ln b}}=c^{\ln a/\ln b}=c^{\log_b a}$.

Answer 2

If you apply the logarithm with base $a$ to both sides you obtain,

$$\log_a a^{\log_b c} = \log_a c^{\log_b a}$$

$$\log_b c = \log_b a \log_a c$$

$$\frac{\log_b c}{\log_b a} = \log_a c$$

however this last equality is the change of base formula and hence is true. Reversing the steps leads to the desired equality.

Answer 3

$a^{\log_b c} =c^{\log_b a}$

let $k= a^{log_b c}$

$\log k=\log( a^{\log_b c})$

$\log k=\log_b c \log a$

$\log k=\frac{\log c}{\log b} \log a$

$\log k=\frac{\log a}{\log b} \log c$

$\log k=\log_b a \log c$

$\log k=\log c^{\log_b a}$

substituting value of k

$\log a^{\log_b c}=\log c^{\log_b a} $

therefore

$a^{\log_b c} =c^{\log_b a}$

Answer 4

In log there is a property that :$x^{\large {\log _x y}}=y$,$\log_x y=\log_w y\times\log_x w$ where $w$ can be any hold any possible value that is valid for a log base. so $$a^{\large{\log_b c}}\implies a^{\large{\log_a c\times \log_b a}}$$ since $a^{\large{\log_a c}}=c$ so $$a^{\large{\log_a c\times \log_b a}}\implies c^{\large{\log_b a}}$$

Answer 5

$a^{\log_b c} = c^{\log_b a}$, needs to be proven

let's assume LHS = x

$a^{\log_b c} = x$

take $log_a$ from both sides

$\log_a{(a^{\log_b c)}} = \log_a(x)$

$\log_b c = \log_a(x)$

change all base to $b$

$\log_b c = \frac{\log_b(x)}{\log_b(a)}$

from this we extract $\log_b(a)$ which is $\log_b a = \frac{\log_b(x)}{\log_b(c)}$

again right side can be written as $\log_c{x}$

so $\log_b a= \log_c{x}$

raise to the power of $c$

$c^{\log_b a}= c^{\log_c{x}}$

$c^{\log_b a}= x$

and we know what $x$ is so

$a^{\log_b c} = c^{\log_b a}$

Answer 6

Has no one hit upon the simplest way to do this, which is to take logarithms to the base $b$?

$\log_b(a^{\log_bc})=\log_bc\log_ba$,
$\log_b(c^{\log_ba})=\log_ba\log_bc$, done!