1. ## [SOLVED] Log Help

Hi,

I need help solving a log equation for x algebraically:

This is the equation: 2log2(x-6) - log2x = 3

Help is much appreciated!This is what I have so far:
2log2(x-6) - log2x = 3 (both logs are base 2)

log2(x-6)^2 - log2x = 3

Im not sure what the next step I should take to solve is?

Any thoughts?

Thanks!

Jim.

2. Hello, Jim!

$2\log_2(x-6) - \log_2x \:= \:3$

This is what I have so far: . $\log_2(x-6)^2 - \log_2x \:= \:3$
Keep going . . .

$\log_2\left[\frac{(x-6)^2}{x}\right] \:=\:3 \quad\Rightarrow\quad \frac{(x-6)^2}{x} \:=\:2^3 \quad\Rightarrow\quad (x-6)^2 \:=\:8x$

And we have: . $x^2-12x + 36 \:=\:8x \quad\Rightarrow\quad x^2 - 20x + 36\:=\:0$

. . which factors: . $(x-2)(x-18) \:=\:0$

. . and has roots: . $x \:=\:2,\:18$

But $x=2$ is an extraneous root.

The only solution is: . $x \,= \,18$

3. Originally Posted by jimstewart_NC
Hi,

I need help solving a log equation for x algebraically:

This is the equation: 2log2(x-6) - log2x = 3

Help is much appreciated!This is what I have so far:
2log2(x-6) - log2x = 3 (both logs are base 2)

log2(x-6)^2 - log2x = 3

Im not sure what the next step I should take to solve is?

Any thoughts?

Thanks!

Jim.
Both sides of the equation contain logarithms. Use the appropriate base to get rid of the logarithms:

$2^{2\log_2(x-6)-\log_(x)} = 2^3$

$\dfrac{(x-6)^2}{x} = 8~\implies~x^2-12x+36=8x$

This is a quadratic equation. Solve for x. I've got: $x = 18~\vee~x = 2$

Since x must be greater or equal 6 the only solution is x = 18

EDIT: Too fast for me - again!