Having trouble solving $\log (x − 21) = 2 − \log x$ for $x$

Question

I'm having trouble with this problem:

$\log (x − 21) = 2 − \log x$, solve for $x$.

I'm coming up with $x=-5$ but that can't be right.

Answer 1

$$\begin{align}\log(x-21) & = 2- \log x \iff e^{\log(x - 21)} = e^{2-\log x}\\ \\ & \iff x-21 = e^2\cdot e^{-\log x}\\ \\ & \iff x-21 = \frac{e^2}{x}\end{align}$$

Can you take it from here? Remember, we must have $x>21$, or else $\log(x- 21)$ will not be defined!

Note: if you're working with $\log_{10}$, then simply replace every $e$ with $10$, and you're good to go! For any base, $\log_b$, you'd use $b$ where I've used $e$.

Answer 2

Assuming that $\log x = \log_{10} x$, here are the steps \[ \log(x-21)=2-\log x \] \[ \log(x-21)+\log x=2 \] \[ \log(x(x-21))=2 \] \[ 10^{\log(x(x-21))}=10^2 \] \[ x(x-21)=x^2-21x=100 \] Now just solve the quadratic equation \[ x^2-21x-100=0 \] \[ (x-25)(x+4)=0 \] Thus the solution is $x=25$. We cannot use $x=-4$ as a solution because $\log(-4)$ is undefined. I hope this helps you understand.