Thread: log help - test tomarrow

1. log help - test tomarrow

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

2. 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: . $\boxed{x \:=\:4}$

3. 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.

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

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