Solving exponential equations using logarithms

Question

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?

Answer 1

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}

Answer 2

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 $100$.

Answer 3

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