Solving for x
log(x^2+9)-2logx=log5
I did:
10^5 = (x^2 + 9) / x^2
x's cancel out so I don't have any x to solve for. So clearly I messed up somewhere
The x's don't cancel out you should have
$\displaystyle 5=\frac{x^2+9}{x^2} \iff \frac{x^2+9}{x^2}-5=0 \iff \frac{x^2+9-5x^2}{x^2}=0 \iff \frac{9-4x^2}{x^2}=0$
This tells us that x cannot be equal to zero and now we need to set the quadratic in the numerator equal to zero and solve. Be sure to check your solutions in the original equaiton!
Hello, Devi09!
$\displaystyle \text{Solve for }x \!:\;\;\log(x^2+9) - 2\log x\:=\:\log5$
Note that $\displaystyle x$ must be positive.
I avoid fractions when possible . . .
We have: .$\displaystyle \log(x^2+9) \:=\:\log 5+ 2\log x$
. . . . . . . . $\displaystyle \log(x^2+9) \:=\:\log5 + \log(x^2) $
. . . . . . . . $\displaystyle \log(x^2+9) \:=\:\log(5x^2) $
"Un-log": . . . . $\displaystyle x^2 + 9 \:=\:5x^2$
And we have: .$\displaystyle 4x^2 \:=\:9 \quad\Rightarrow\quad x^2 \:=\:\tfrac{9}{4} \quad\Rightarrow\quad x \:=\:\pm\tfrac{3}{2}$
Therefore: .$\displaystyle x \:=\:\frac{3}{2}$