solve $ \log_2 x= 3- 2 \log_x 2$

Question

note: space= base and logarithm difference. no space= multiply

My method

$\log_2x=3-2\log_x2$

$\log_ x 2=\log_ 2 x $

$\frac{1}{\log_2 x}= 3-2\log _x 2$

$\frac{1}{\log_2 x}= \log_x (x^3/4)$

$\frac{1}{\log_2 x}= \frac{\log_2 (x^3/4)}{\log_2 x}$

$1 = \log_2 4x^3$

$x^3 = 8$

$x = 2$

Ideal answer: Substitute $u=\log_2 x$

$u = 3-2/u$

$u^2-3u+2 = o$

$u=1,2$

$x=2$ and $ x=4$

Please help me and explain to me what is wrong in my equation that I am unable to find the second value of $x$

Re: I finally noticed the $\log_2 x$ into $\frac{1}{\log_2 x}$ substitution, so I am sorry that I'm asking a useless question. its merely coincidental that I have found 1 value of $x$ with wrong workings.

Answer 1

The first step seems wrong

$$\log_2x=3-2\log_x2\implies \color{red}{\frac{1}{\log_2 x}}= 3-2\log_x 2$$

indeed it should be

$$\log_2x=3-2\log_x2\implies \log_2x= 3-2\frac{1}{\log_2 x}$$

Answer 2

Note: unless named, $\log$ is in base 10.

Given: $\log_2x=3-2\log_x2$, then $$\log_2x=3-\log_x2^2=3-\log_x4$$ $$\frac{\log x}{\log 2}=3-\frac{\log4}{\log x}\Rightarrow\frac{\log x}{\log 2}+\frac{\log4}{\log x}=3$$ $$\frac{(\log x)^2+\log 4\log2}{\log x\log2}=3\Rightarrow(\log x)^2+\log 4\log2=3\log x\log2$$ $$(\log x)^2+\log 4\log2-3\log x\log2=0$$ Let $q=\log x\Rightarrow q^2-(3\log2)q+\log4\log2$. Which can be solved using the quadratic formula: $$q=\frac{3\log2\pm\sqrt{(-3\log2)^2-4(1)(\log4\log2)}}2$$

To solve for $x\Rightarrow\log x=\frac{3\log2\pm\sqrt{(-3\log2)^2-4(1)(\log4\log2)}}2$ $$x=\exp \left(\frac{1}{2} \left(3 \log (2)+\sqrt{\log (2) (9 \log (2)-4 \log (4))}\right)\right)$$ $$x=\exp \left(\frac{1}{2} \left(3 \log (2)-\sqrt{\log (2) (9 \log (2)-4 \log (4))}\right)\right)$$ $$\bbox[yellow,10px]{\therefore x=4.,x=2.}$$