Prove $a^{\log_bc}=c^{\log_ba}$

Question

Can anyone prove $$a^{\log_bc}=c^{\log_ba}$$ I've tried using algebra but I always get $a=a$ $c=c$ or $b=b$. Sometimes I get the same property but now flipped. Can anyone help?

Answer 1

taking the natural logarithm on both sides we get $$\log_{b}{c}\ln(a)=\log_{b}{a}\ln(c)$$ and this can be written as $$\frac{\ln(c)\ln(a)}{\ln(b)}=\frac{\ln(a)\ln(c)}{\ln(b)}$$ which is the same

Answer 2

Hint:

Take $\log_b$ on both sides.

Answer 3

Take $\log_b$ of both sides: $$a^{\log_bc}=c^{\log_ba}$$ $$\iff\log_ba^{\log_bc}=\log_bc^{\log_ba}$$ $$\iff\log_bc\log_ba=\log_ba\log_bc$$ This last equality is trivially true by commutativity of multiplication.

Answer 4

Start with LHS, let $y=a^{\log_b c}$

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

Thus, $\log_b y = \log_b c^{\log_b a}\implies y = c^{\log_b a}$ and so

$$a^{\log_b c}=y=c^{\log_b a}$$ LHS = RHS

Answer 5

$$a^{\log_bc}=b^{\log_ba\log_bc}=\left(b^{\log_bc}\right)^{\log_ba}=c^{\log_ba}.$$

Answer 6

$$ \large{a^{\log_bc}=c^{\log_ba}}$$

Take logs on either side to any real base

$${\log_bc} \log a = {\log_ba} \log c $$

again take logs either side to any real base for first terms each side

$$ \dfrac{\log c \log a }{\log b}= \dfrac{\log a \log c }{\log b}$$

and it tallies.

Btw, we can swap $(a,c)$ in the given equation.

Answer 7

For fun:

Start with the identity:

$\log_b(c) \cdot \log_b(a) = \log_b(c)\cdot \log_b(a);$

Take $\exp_b$ of both sides:

$\exp_b(\log_b(c) \log_b(a))=$

$ \exp_b(\log_b(a)\log_b(c))$;

$\exp_b(\log_b(a)^{\log_b(c)})=$

$\exp_b(\log_b(c)^{\log_b(a)}).$

Hence :

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