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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.

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  • $\begingroup$ $\log_b(a+c)\ne\log_ba+\log_bc$ $\endgroup$
    – Aiden Chow
    Jan 19, 2022 at 0:33
  • $\begingroup$ 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. $\endgroup$
    – Frederick
    Jan 19, 2022 at 0:46
  • $\begingroup$ @Frederick Please see the answer I've just posted. $\endgroup$
    – Aiden Chow
    Jan 19, 2022 at 0:53

2 Answers 2

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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.

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  • $\begingroup$ Aiden, thank you very much. Your explanation helped me greatly. Thanks again! $\endgroup$
    – Frederick
    Jan 19, 2022 at 1:03
  • $\begingroup$ @Frederick If you feel that your question has been answered to your satisfaction, please click the green check mark next to the answer. Thanks! $\endgroup$
    – Aiden Chow
    Jan 19, 2022 at 1:08
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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?

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  • $\begingroup$ Si. Muchas Gracias por la Ayuda. Buena Suerte con Su Ph.D.Candidatura. $\endgroup$
    – Frederick
    Jan 20, 2022 at 2:08
  • $\begingroup$ @Frederick Muchas gracias ! :D $\endgroup$ Jan 20, 2022 at 2:14

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