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This is the equation that I am having troubles with:

$$\large x^{\large\log_{10}5}+5^{\large\log_{10}x}=50$$

So the first thing I do, I logarithm the whole expression with $\log_{10}$.
So I get:

$ {\log_{10} 5} \times {\log_{10} x} + {\log_{10} 5} \times {\log_{10} x} = {\log_{10} 50}$

When I solve this one for $x$, I get that $x = 16$, which is totally incorrect because it is supposed to be $100$. Can anyone tell me what am I doing wrong or show me how to solve this equation?

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Let $y=x^{\large\log_{10}5}$, then $$ \log_{10}y=(\log_{10}5)(\log_{10}x)=\log_{10}5^{\large\log_{10}x}\color{red}{\quad\Rightarrow\quad} y=5^{\large\log_{10}x}. $$ Hence \begin{align} x^{\large\log_{10}5}+5^{\large\log_{10}x}&=50\\ 5^{\large\log_{10}x}+5^{\large\log_{10}x}&=50\\ 2\times5^{\large\log_{10}x}&=50\\ 5^{\large\log_{10}x}&=25\\ 5^{\large\log_{10}x}&=5^2\color{red}{\quad\Rightarrow\quad}\log_{10}x=2\color{red}{\quad\Rightarrow\quad}\large\color{blue}{ x=10^2=100}. \end{align}

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  • $\begingroup$ Thanks a lot, this really helped :) $\endgroup$ – Dimitar Spasovski May 28 '14 at 11:08
  • $\begingroup$ You are welcome @Kockar :) $\endgroup$ – Tunk-Fey May 28 '14 at 11:11
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Your first action was bad, because the identity you assumed ("$\log{a+b}=\log{a}+\log{b}$") simply doesn't exist.

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  • $\begingroup$ I have always liked the equation : $\log (1+2+3)=\log 1+\log 2 +\log 3$ $\endgroup$ – evil999man May 28 '14 at 12:51
  • $\begingroup$ @Awesome It is the same as $\log{ab}=\log{a}+\log{b}$ :-) $\endgroup$ – peterh May 28 '14 at 12:56
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If $x > 100$, then: $LHS \gt 100^{log_{10}^5} + 5^{log_{10}^{100}} = \left(10^{log_{10}^5}\right)^2 + 5^2 = 5^2 + 5^2 = 50 = RHS$,

similarly if $x < 100$, then $LHS < RHS$. Thus $x$ can only be $100$.

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  • $\begingroup$ But how do I get that 100 so I can start calculating and comparing? $\endgroup$ – Dimitar Spasovski May 28 '14 at 10:50
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    $\begingroup$ It can be done without using trial & error like Tunk-Fey's answer $\endgroup$ – Anastasiya-Romanova 秀 May 28 '14 at 10:59
  • $\begingroup$ OK, fine. Whatever... $\endgroup$ – Anastasiya-Romanova 秀 May 28 '14 at 11:18

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