Is there any way to solve $\log^2_{10}(x) + \log_{10}(x)\log_{10}(2) - \log_{10}(5)=0$?

Question

I am asking how to solve this without using a calculator. The original question is:

Find $x$ if $x^{\log_{10}(2x)} = 5$.

I started off by writing $5$ and $x$ as $10^{\log_{10}(5)}$, $10^{\log_{10}(x)}$.

This way we will have $\log_{10}(x)\log_{10}(2x)=\log_{10}(5)$ which is equivalent to $\log_{10}(x)\log_{10}(2x)-\log_{10}(5)=0$.

Writing $\log_{10}(2x)$ as $\log_{10}(2) + \log_{10}(x)$ gives exactly what I had in the title: $$\log^2_{10}(x) + \log_{10}(x)\log_{10}(2) - \log_{10}(5)=0$$

Discriminant (don't know the English word for it) is equal to $\log^2_{10}(2) + 4\log_{10}(5)$, which you can write as $\log^2_{10}(2) + \log_{10}(5^4)$, I don't know which one helps more.

Let's think of $\log_{10}(x)$ as a real number $t$. We would then have:

$$t = \frac{-\log_{10}(2) \pm \sqrt{\log^2_{10}(2) + 4\log_{10}(5)}}{2}$$ Using a calculator we can see that the solutions of this equation are $t = -1, \log_{10}(5)$ therefore $x = \frac{1}{10}, 5$.

I am asking if there is any possible way to solve this problem without a calculator/AI using the method I have written above.

If you know the answer to my question or have an easier solution/hints for an easier one, I would appreciate your help. Thanks in advance!

Answer 1

I believe the only puzzle piece you're missing is $$ \log_{10}(2) + \log_{10}(5) = \log_{10}(2\times5) = \log_{10}(10) = 1, $$ which means that $\log_{10}(5) = 1 - \log_{10}(2)$. With that identity you should be able to simplify the roots you got from the quadratic equation by hand.