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.
$\endgroup$ 32 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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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