Solving a logarithmic equation where the logarhitm is exponentiated

Question

I have troubles solving the following logarthitmic equation.

$$ \ 2(\log_x{\sqrt7})^2-\log_x{\sqrt7}-1 =0 $$

The results are supposed to be $ \ x_1 = {\frac{1}{7}}, x_2 = \sqrt7 $

I have tried substitution ( $\log_x{\sqrt7} = x)$, which yielded a quadratic equation with results of $ { x_1 =\frac{1}{2} } $ and $ x_2 = 0 $, none of which works. I would like to ask what the proper way of solving equations like this is?

Answer 1

The key is to substitute $\log_x 7$ with a separate variable (not $x$) and then solve the quadratic

$$ \log_x \sqrt{7} = \frac{1}{2}\log_x 7$$

let $\log_x 7 = a$

Your equation reduces to:

$$2\left(\frac{1}{2}\log_x 7\right)^2 -\frac{1}{2}\log_x 7 - 1 = 0$$

$$ a^2 -a -2 = 0$$

Solve the quadratic and then solve for $x$.

Answer 2

Hint

$\log_x\sqrt{7}$ is a number.

$y$ is a number.

$\gamma$ is a number.

Our equation states that twice a number squared less the number less one is equal to zero.

or $$2\gamma^2-\gamma-1=0$$ or $$2y^2-y-1=0$$ or alternatively $$2(\log_x\sqrt{7})^2-(\log\sqrt{7})-1=0$$