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I have an easy problem. I can see the answer but I don't know how to solve the problem and get the answer "the mathematical way".

The statement is: $x \large {\cdot 2^{\log _x 5 } = 10 }$

Then I brought it to the form $\log _x 25^x = 10$

The answer is clearly 5, but how do I prove it mathematically?

Andrew
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    \large x \cdot 2^{\log _x 5 } = 10 to make it look bigger, like this $\large x \cdot 2^{\log _x 5 } = 10$ – Git Gud Feb 03 '13 at 11:53

1 Answers1

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Using $\log_a b=\frac{\ln b}{\ln a}$ and logarithm rules we can take logarithm of the given equation and find $$\begin{array}\ &x\cdot 2^{\log_x5}=10\\ \Rightarrow&\ln x+\log_x5\cdot\ln 2=\ln 10\\ \Rightarrow&\ln x+\frac{\ln5}{\ln x}\ln 2=\ln 10\\ \Rightarrow&(\ln x)^2-\ln10\cdot \ln x+{\ln5}\ln 2=0\\ \Rightarrow&(\ln x)^2-(\ln5+\ln2)\cdot \ln x+{\ln5}\ln 2=0\\ \Rightarrow&(\ln x-\ln5)(\ln x-\ln 2)=0\\ \Rightarrow&\ln x-\ln5=0\quad\text{or}\quad\ln x-\ln 2=0\\ \Rightarrow&x=5\quad\text{or}\quad x=2.\\ \end{array}$$ And in fact $2\cdot 2^{\log_2 5}=2\cdot 5=10$ and $5\cdot 2^{\log_5 5}=5\cdot 2^1=10$.

Hagen von Eitzen
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