How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$? In one of my recent exam, I was ask to solve this:
$$
10^{x^2+x}+\log{x} = 10^{x+1}
$$
My attempt to solve it was:
$$
10^{x^2+x}+\log{x}       = 10^{x+1}         \\
\log{x}=10^{x+1}-10^{x^2+x}  \\
\log{x} = 10^{x+1}(1-10^x)   \\
\log(\log{x})=(x+1)+\log(1-10^x)    \\
$$
At this point I got stuck because I don't know how to solve an equation with double logs.

If $x > 1$, then $x^2 + x > x + 1> 0$, and $\log x > 0$, thus $LHS > RHS$, and if $0 < x < 1$, then $0 <x^2 + x < x + 1$, and $\log x < 0$, thus $LHS < RHS$. When $x = 1$, both sides equal to $100$. Thus the only answer is $x = 1$.
