Help with $\frac12 \log_2 x - \frac1{\log_2 x} = \frac76$

Question

I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?

Answer 1

Note that

$$\left( \log_2(x) \right)^{-1} \neq \log_2 \left(x^{-1} \right)$$

For example, take $x = 4$. Then

$$\left( \log_2(x) \right)^{-1} = \left( \log_2(4) \right)^{-1} = 2^{-1} = \frac 1 2$$

but

$$\log_2 \left(x^{-1} \right) = \log_2 \left( \frac 1 4 \right) = -2$$

This is where your error lies.

Answer 2

To get the correct answer, let $L=\log_2(x).$

Then we have $$\frac 1 2 L - \frac 1 L = \frac 7 6.$$

Multiply by $6L$ to get $$3L^2-6=7L.$$

Thus $$3L^2-7L-6=0$$

or $$(3L+2)(L-3)=0.$$

Can you take it from here?