How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$

Question

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.

Answer 1

There is a mistake later as well, but here's the first, from the fourth to fifth line: $$ \boxed{ \log \left(\frac x{10}\right) \neq \frac{\log_x x}{\log_x {10}}} $$

Because the left hand side is $\log_{10} x - 1 = \frac{1}{\log_{x}10} - 1$ while the RHS is $\frac 1{\log_x 10}$ , and these two are clearly not equal, in fact they differ by one.

Thanks to this a second error, namely $(\log_x 10)^2 \neq \log_x (10^2)$ becomes inconsequential(this is the transition from sixth to seventh line).

Correcting that to $\log\left(\frac{x}{10}\right) = \log_{10} x - 1$ gives $$ \frac{-1}{4} \log_{x} 10 = \log_{10} x - 1 $$

now use the fact that $\log_{10}x \log_{x}10 = 1$ to see that if $y = \log_{x} 10$ then $-y = \frac 4y - 4$, etc. , it is a quadratic equation which you can solve easily to get the desired $y,x$ values.

Answer 2

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