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.
 A: Note that $x=1$ is obviously a solution. We prove that this is the only solution.
Suppose that $x>1$. Then $x^2>1$, so $x^2+x>x+1$. Exponentiating both sides we get that:
$$10^{x^2+x}>10^{x+1}$$
since $\log(x)>0$ when $x>1$, then we also have:
$$10^{x^2+x}+\log(x)>10^{x+1}$$
so there are no solutions in this range.
Similarly, when $0<x<1$, then $x^2+x<x+1$, so
$$10^{x^2+x}<10^{x+1}$$
and because $\log(x)<0$ in this range, then:
$$10^{x^2+x}+\log(x)<10^{x+1}$$
so there are no solutions in this range either.
The equation is undefined when $x\leq0$, so this proves that $x=1$ is the only solution.
A: 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 $x = 1$.
A: In your manipulation of the equation you factored the powers of $10$ incorrectly. Correction:
$$10^{x^2+x} + \log(x) = 10^{x+1}\\\log(x) = 10^{x+1}(1-10^{x^2 - 1})$$

We know that $x > 0$ for $\log(x)$ to be defined.
Case ($0 < x < 1$): $\log(x) < 0 < 10^{x+1}(1-10^{x^2-1})$
Case ($x = 1$): $\log(1) = 0 = 10^2(1-10^0)$
Case ($x > 1$): $\log(x) > 0 > 10^{x+1}(1-10^{x^2-1})$
Thus the only solution is $\boxed{x = 1}$
