Solve this equation for $x$ Let $$ax^{\log(a)}=bx^{\log(b)}$$ Solve this equation for $x$ when $a\neq b $.
My work thus far: $$ax^{\log(a)}=bx^{\log(b)}$$ $$\log(ax^{\log(a)})=\log(bx^{\log(b)})$$ $$\log(a)\log(ax)=\log(b)\log(bx)$$ $$\log(a)[(\log(a)+\log(x))]=\log(b)[(\log(b)+\log(x))]$$ $$\log^2(a)+\log(a)\log(x)=\log^2(b)+\log(b)\log(x)$$ $$\log^2(a)-\log^2(b)=\log(b)\log(x)-\log(a)\log(x)$$ $$\log^2(a)-\log^2(b)=\log(x)[(\log(b)-\log(a))]$$ $$\frac{\log^2(a)-\log^2(b)}{\log(b)-\log(a)}=\log(x)$$ $$x=e^{\frac{\log^2(a)-\log^2(b)}{\log(b)-\log(a)}}=exp(\frac{\log^2(a)-\log^2(b)}{\log(b)-\log(a)})$$
Does this seem correct? Upon referencing my answer against Wolfram, I get this answer ($x=\frac{1}{e}$). Is there any way to simplify this answer? Thanks in advance. 
 A: $$ax^{\ln a}=bx^{\ln b}$$
Taking logarithm base $e,$
$$\ln a+\ln (x^{\ln a})=\ln b+\ln (x^{\ln b})$$
$$\iff\ln a+\ln a\cdot\ln x=\ln b+\ln b\cdot\ln x$$
$$\iff\ln x=\dfrac{\ln a-\ln b}{\ln b-\ln a}=-1$$ assuming $\ln a\ne\ln b$ 
A: $$\frac{a}{b}=\frac{x^{\log b}}{x^{\log a}}$$
$$\frac{a}{b}=x^{\log b - \log a}$$
$$\frac{a}{b}=x^{\log \frac{b}{a}}$$
$$\sqrt[log\frac{b}{a}]{\frac{a}{b}}=x$$
A: Your mistake is in the third line $$\log(a)\log(ax)=\log(b)\log(bx)$$ should be $$\log a+\log(ax)= \log b + \log(bx)$$
A: Frankly I'd just do this:
$ax^{\log a} = bx^{\log b}$
$\frac ab  =  x^{\log b - \log a} = x^{\log \frac ba}$
$\log \frac ab = \log \frac ba * \log x= - \log \frac ab \log x$
$\log x = -1$
$x = \frac 1{e}$
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Or more directly:
$ax^{\log a} = ae^{\log x \log a}$ and $bx^{\log b} = be^{\log x \log b}$
So $\frac ab = e^{\log x (\log b - \log a)}= e^{\log x \log \frac ba} = (\frac ba)^{\log x}$
So $\log_{\frac ba} \frac ab = \log_{\frac ba} (\frac ba)^{\log x}$
$-1 = \log x$.
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Or $ax^{\log a} = ae^{\log a\log x} = a*a^{\log x} = a^{1 + \log x}$.
And $bx^{\log b} = b^{1 + \log x}=a^{1 + \log x}$
So $(\frac ab)^{1+\log x} = 1$.
So $1 + \log x = \log_{\frac ab} 1 = 0$
So $\log x = -1$.
Or
.......
A: From saulspatz's answer, you should get:
$$\begin{align}
\ln a+\ln x^{\ln a}&=\ln b+\ln x^{\ln b}\\
\ln a+\ln a (\ln x)&=\ln b+\ln b(\ln x)
\end{align}$$
Skipping a few steps, you must get lab bhattacharjee's answer:
$$\begin{align}
\ln x=\frac{\ln b-\ln a}{\ln a-\ln b}\Rightarrow\frac{\ln \frac ba}{\ln \frac ab}\\
\end{align}$$
And by the change of base rule, you get:
$$\begin{align}
\ln x=\frac{\ln b-\ln a}{\ln a-\ln b}&\Rightarrow\frac{\ln \frac ba}{\ln \frac ab}\\
&\Rightarrow\log_{\frac ab}\frac ba\Rightarrow-1
\end{align}$$
That's why you get:
$$$$\begin{align}
\ln x=-1\Rightarrow x=\frac1e
\end{align}$$$$
