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

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!

• "Discriminant" is exactly correct in English! Commented Feb 17, 2022 at 17:44

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.