Solving exponential equations using logarithms

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?

-

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}

-
Thanks a lot, this really helped :) – Dvorog May 28 '14 at 11:08
You are welcome @Kockar :) – Tunk-Fey May 28 '14 at 11:11

Your first action was bad, because the identity you assumed ("$\log{a+b}=\log{a}+\log{b}$") simply doesn't exist.

-
I have always liked the equation : $\log (1+2+3)=\log 1+\log 2 +\log 3$ – evil999man May 28 '14 at 12:51
@Awesome It is the same as $\log{ab}=\log{a}+\log{b}$ :-) – peterh May 28 '14 at 12:56

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 $100$.

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