Logarithm Equation - Lost What to do with a squared logarithm?

Question

Struggling with a logarithm problem here:

Solve the quation for x:

$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$

My initial thought was to present the equation in such a way as to get rid of the 'x' base and use the exponents solemnly. I.e.

$x ^{2 - (lgx)^2-lgx^2}-1/x = 0$

Becomes: $x ^{2 - (lgx)^2-lgx^2}-x^{-1} = 0$ Which leads to: $2-(lgx)^{2} -lgx^{2} +1 = 0 $

However, from here on, I am absolutely lost in regards to what to do. I have no clue what to do with the squared logarithm, and in turn, no idea how to solve this.

The answers given by the textbook are: x1=1 ; x2=10 ; x3=0.001

I must admit, I'm even double lost as to how to reach 3 answers, haha.

I am preparing for my math exam (and I am studying on my own, no tutor) so any help would be much appreciated!

Thank you in advance and happy holiday preparations! :)

Answer 1

Note that $x=1$ is always a solution since $1$ raised to any power is $1=1/x$.

Take $\log x=A$, you get $(\log x)^2+\log(x^2)=(\log x)^2+2\log(x)=A^2+2A=3$

$\displaystyle\implies(A+3)(A-1)=0, A=-3,1\\\therefore x=10^{-3},10$

Answer 2

Hint: clearly $x \ne 0$ so multiplying by $x$ we get $x ^{3 - (\lg x)^2-\lg x^2} = 1$. So either $x=1$ or $3 - (\lg x)^2-\lg x^2=0$. The last equation you can solve by taking $y=\lg x$ and solving $3-y^2-2y=0$.

Answer 3

Recall that $\log(x^2) = 2\log x$; so, you have to solve $$ 3 - (\log x)^2 - 2\log x = 0 $$ for $x>0$. Set $y = \log x$. Then, the original equation is equivalent to solving $$ 3 - y^2 -2y =0 $$ (and then, given a solution $y^\ast$, $x^\ast = e^{y^\ast}$ will be a solution of the original equation). But this is now a quadratic equation, which you know how to handle.

Important: when you "got rid of $x$ in the base", you implicitly simplified by $\log x$. So you may have discarded a solution, $x=1$ (for which $\log x=0$). You have to check separately whether this was a solution of the original equation.