log help - test tomarrow

**Porcelain** · April 2nd 2009, 04:40 PM

im not sure how to type logs so try to bare with me >< first set of parenthesis is the base of the log
log(4)x + log(4)(x-2)= log(4)8
if you could explain that i would be super happy

**Soroban** · April 2nd 2009, 04:49 PM

Hello, Porcelain!

$\log_4(x) + \log_4(x-2) \:=\:\log_4(8)$

On the left side, combine the logs . . .

We have: . $\log_4\bigg[x(x-2)\bigg] \:=\:\log_4(8)$

"Un-log" both sides: . $x(x-2) \:=\:8 \quad\Rightarrow\quad x^2 - 2x - 8 \:=\:0$

. . Then: . $(x-4)(x+2)\:=\:0 \quad\Rightarrow\quad x \:=\:4,-2$

But $x = -2$ is an extraneous root.

. . Therefore, the solution is: . $\boxed{x \:=\:4}$

**skeeter** · April 2nd 2009, 04:51 PM

Originally Posted by Porcelain

im not sure how to type logs so try to bare with me>< first set of parenthesis is the base of the log
log(4)x + log(4)(x-2)= log(4)8
if you could explain that i would be super happy

$\log_4{x} + \log_4(x-2) = \log_4{8}$

$\log_4[x(x-2)] = \log_4{8}$

$x(x-2) = 8$

solve the quadratic ... don't forget to check both solutions in the original equation for domain issues.

**Porcelain** · April 2nd 2009, 04:57 PM

so if a log of the same base is equal to a log of the same base you get rid of the logs?

**daunder** · April 2nd 2009, 04:58 PM

Log a + Log b = Log (a*b) and yes you get rid of the logs if they have same base

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