Prove that $a^{\ln{(b)}} = b^{\ln{(a)}}$

Question

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

They don't have to be different from 1. If you check, subbing 1 for either constant works. It's just not interesting. — Kaynex, Nov 05 '16 at 23:51
The "solution" doesn't prove the claim for $a=b=1$, true. The problematic step is when it takes $\log_a$ both sides when $\log_a$ cannot be defined for $a=1$. — user160738, Nov 05 '16 at 23:52
Your book gives a too complicated solution. Instead, take ln of both sides. Thus there is no problem with $a=1$ and $b=1.$ — Jean Marie, Nov 05 '16 at 23:54
@AliceIsDead $\log_a (x)$ is defined as the inverse function of $a^x$, but when $a=1$ the function $a^x$ is identically $1$, so it is far from being bijective. So inverse cannot exist — user160738, Nov 05 '16 at 23:55
What it has done, it has used the fact that $\log_{a} (b) = \frac{log_{c}(a)}{\log_{c} (b)}$ (for any positive $a,b,c$). For an easier proof you can do this $a^{\ln (b)} = e^{\ln(a) \ln(b)}$, and the same for the right side. — kolobokish, Nov 05 '16 at 23:57
The issue does not exist $1^{\ln(1)}=1^0=1$ is perfectly defined ! — Jean Marie, Nov 05 '16 at 23:57
@AliceIs Dead we don't have to assume $a=b$. It is true for any $a,b>0$. — Jean Marie, Nov 05 '16 at 23:59
Possible duplicate of Logarithm proof problem: $a^{\log_b c} = c^{\log_b a}$ — Mohan Sharma, Jan 24 '18 at 08:38

Bernard · Answer 1 · 2016-11-06T00:05:17.617

8

This is anyway a complicated solution. Just use the definition of $x^y$ for $x>0$: $$\begin{cases} a^{\ln b}=\mathrm e^{\ln b\,\ln a}, \\ b^ {\ln a}=\mathrm e^{\ln a\,\ln b}. \end{cases}$$

edited Nov 06 '16 at 00:05

answered Nov 05 '16 at 23:57

Bernard

175,478

score 4 · Accepted Answer · answered Nov 05 '16 at 23:57

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.

Prove that $a^{\ln{(b)}} = b^{\ln{(a)}}$

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