Solving x from log equation

**Devi09** · May 11th 2011, 08:17 AM

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

**TheEmptySet** · May 11th 2011, 08:22 AM

Originally Posted by Devi09

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

$5=\frac{x^2+9}{x^2} \iff \frac{x^2+9}{x^2}-5=0 \iff \frac{x^2+9-5x^2}{x^2}=0 \iff \frac{9-4x^2}{x^2}=0$

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!

**Soroban** · May 11th 2011, 09:23 AM

Hello, Devi09!

$\text{Solve for }x \!:\;\;\log(x^2+9) - 2\log x\:=\:\log5$

Note that $x$ must be positive.

I avoid fractions when possible . . .

We have: . $\log(x^2+9) \:=\:\log 5+ 2\log x$

. . . . . . . . $\log(x^2+9) \:=\:\log5 + \log(x^2)$

. . . . . . . . $\log(x^2+9) \:=\:\log(5x^2)$

"Un-log": . . . . $x^2 + 9 \:=\:5x^2$

And we have: . $4x^2 \:=\:9 \quad\Rightarrow\quad x^2 \:=\:\tfrac{9}{4} \quad\Rightarrow\quad x \:=\:\pm\tfrac{3}{2}$

Therefore: . $x \:=\:\frac{3}{2}$

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