<s>[INST] 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.
 [/INST] 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 $\boxed{x = 1}$
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