How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$?

Question

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.

Answer 1

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 $x = 1$.

Answer 2

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.

Answer 3

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 $\boxed{x = 1}$