What is the solution to $\log_{10} x -x=2?$ What is the solution to $\log_{10} x -x=2?$
I have tried to solve it but I couldn't. I've got to $x^x =200$.
 A: Solving $\log_{10}x - x = 2$ is equivalent to solving $x = 10^{x+2}$. Now note that $x = 0 < 10^{0+2}$ and that the function $10^{x+2}$ grows faster than the function $x$ for $x \geq 0$. From this we can conclude that $x < 10^{x+2}$ for all $x \geq 0$. Therefore, the equality can't be valid.
Another way to see this is to plot the functions.
A: $f(x)=log_{10}(x)-x$ remains completely within the fourth quadrant, so $f(x)$ will never be equal to $2$. 
A: Assuming you're using the principal branch of log, 
$$x = -\frac{W(-100 \ln(10))}{\ln(10)} $$
where $W$ is the principal branch of the Lambert W function.  Other branches of log would correspond to other branches of Lambert W.  Since $-100 \ln(10) < -1/e$, no solutions are real.
A: If $x=10^{x+2}$ has a real solution  such solution has to be $\geq 0$, since $10^{x+2}\geq 0$ for any $x\in\mathbb{R}$.
But if $x\geq 0$ and $x=10^{x+2}$ then $x\geq 100$ since $10^{x+2}$ is an increasing function.
If $x\geq 100$ and $x=10^{x+2}$ then $x\geq 10^{102}\geq 10^{10^2}$ for the same reason.
At the next step we get $x\geq 10^{10^{10^2}}$ and by iterating this argument it should be clear that there are no real solutions.
