# What is wrong with my approach to solving $x^{\log25} + 25^{\log x} = 10\;$?

Found this equation on the web: $$x^{\log25} + 25^{\log x} = 10$$ The person solved by substitution and got $$x = \sqrt{10}$$ which satisfies the equation.

I tried different ways after following the man's substitution method. I tried this: $$\log\left(x^{\log25}\right) + \log\left(25^{\log x}\right) = \log 10$$

Using laws of logs: \begin{align} \log25 \cdot \log x + \log x\cdot\log 25 &= \log 10\\ 1.3979 \log x + \log x (1.3979) &= 1\\ 2.7958 \log x &= 1\\ \log x &= 0.357\\ x &= 10^{0.3576} \approx 2.278 \end{align} This is wrong, however.

Why aren't the laws of logs holding? I'm missing something.

Sorry for asking what I'm sure is a ignorant question. I like math but am hardly an expert.

• $\log_b(a+c)\ne\log_ba+\log_bc$ Jan 19 at 0:33
• Thank you, Aiden. I'm sorry but I don't see where this applies to what I did. Could you elaborate a bit. Thank you again. Not a math wizard. Jan 19 at 0:46
• @Frederick Please see the answer I've just posted. Jan 19 at 0:53

The problem with your approach is your first step: taking the log of both sides. When you take the log of both sides, you get $$\log(x^{\log25}+25^{\log x})=\log10$$ This is valid. However, you then somehow transform this equation into what you have written in your question: $$\log(x^{\log25})+\log(25^{\log x})=\log10$$ This is now invalid, because of the fact that $$\log_b(a+c)\ne\log_ba+\log_bc$$. That is to say, you can't simply split a sum inside a logarithm like what you did.

• Aiden, thank you very much. Your explanation helped me greatly. Thanks again! Jan 19 at 1:03
• @Frederick If you feel that your question has been answered to your satisfaction, please click the green check mark next to the answer. Thanks! Jan 19 at 1:08

HINT

Here is a solution with which you can compare:

\begin{align*} x^{\log(25)} + 25^{\log(x)} = 10 & \Longleftrightarrow 10^{\log(x^{\log(25)})} + 10^{\log(25^{\log(x)})} = 10\\\\ & \Longleftrightarrow 2\times 10^{\log(25)\log(x)} = 10\\\\ & \Longleftrightarrow 10^{\log(25)\log(x)} = 5\\\\ & \Longleftrightarrow \log(25)\log(x) = \log(5)\\\\ & \Longleftrightarrow \log(x) = \frac{\log(5)}{\log(25)}\\\\ & \Longleftrightarrow \log(x) = \frac{1}{2} \end{align*}

Can you take it from here?

• Si. Muchas Gracias por la Ayuda. Buena Suerte con Su Ph.D.Candidatura. Jan 20 at 2:08
• @Frederick Muchas gracias ! :D Jan 20 at 2:14