How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$ Question: 
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
My attempts to solve:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
$x^{\log(x)-\log(10) } = 10^\frac{-1}{4}$
$x^{\log(\frac{x}{10}) } = 10^\frac{-1}{4}$
$-\frac{1}{4}\log{_x 10} = \log(\frac{x}{10})$
$-\frac{1}{4}\log{_x 10} = \frac{\log{_x}x}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10} = \frac{1}{\log{_x}10}$
$-\frac{1}{4}\log{_x 10^2} = 1$
$-\frac{1}{2}\log{_x 10} = 1$
$\log{_x 10} = -2$
$10 = x^{-2}$
$10 = \frac{1}{x^2}$
$x^2 = \frac{1}{10}$
$x = {\sqrt{\frac{1}{10}}}$
But this is still wrong. the correct answer is $x = {\sqrt{10}}$. Can someone please point out which part i have done wrong? Thanks. 

$$ x^{\log_{10} x-1}=10^{-1/4} \implies 10^{\log_{10} (x^{\log_{10} x-1})}=10^{1/4} $$ $$\implies 
10^{\log^2_{10} x -\log_{10} x-1/4}=10^0$$ $$\implies (\log_{10} x-1/2)^2=0 \implies \log_{10}x=1/2 \implies x=10^{1/2}=\sqrt{10}.$$
