Prove $a^{\log_bc}=c^{\log_ba}$ 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?
 A: 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.
A: Hint:
Take $\log_b$ on both sides.
A: 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
A: 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
A: $$a^{\log_bc}=b^{\log_ba\log_bc}=\left(b^{\log_bc}\right)^{\log_ba}=c^{\log_ba}.$$
A: 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 :
$a^{\log_b(c)}= c^{\log_b(a)}$
A: $$ \large{a^{\log_bc}=c^{\log_ba}}$$
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.
