Prove that $a^{\ln{(b)}} = b^{\ln{(a)}}$
Prove that $a^{\ln{(b)}} = b^{\ln{(a)}}$ for any values of $a$ and $b$ that are positive and different from 1.
My book says the solution is: \begin{align*} \log_a(a^{\ln(b)})=\log_a(b^{\ln(a)}) &\Leftrightarrow \ln(b) \cdot \log_a{a} = \ln(a)\cdot \log_a(b) \\ &\Leftrightarrow \frac{\ln(b)}{\ln(a)} = \log_ab \\ &\Leftrightarrow \log_ab = \log_ab \end{align*}
I can understand this solution and I understand that since this is a logarithm, all the input values (a and b) must be positive. What I don't understand is how this proves that a and b have to be different from one. Can anyone explain that to me?
Thanks
$\endgroup$ 112 Answers
$\begingroup$This is anyway a complicated solution. Just use the definition of $x^y$ for $x>0$: $$\begin{cases} a^{\ln b}=\mathrm e^{\ln b\,\ln a}, \\ b^ {\ln a}=\mathrm e^{\ln a\,\ln b}. \end{cases}$$
$\endgroup$ $\begingroup$Clearly for positive $a,b$ one has $$ \log(a)\log(b)=\log(b)\log(a). $$ Using "the $x\log(y)=\log(y^x)$-property" of logarithms you have that $$ \log(b^{\log(a)})=\log(a^{\log(b)}). $$ As the logarithm is an injective function this implies $$ b^{\log(a)}=a^{\log(b)}. $$ There is no need to assume that $a\neq1$ or $b\neq1$, it is just that that case is obvious.
$\endgroup$
