Hello, Jim!

Keep going . . .

This is what I have so far: .

And we have: .

. . which factors: .

. . and has roots: .

But is an extraneous root.

The only solution is: .

Results 1 to 3 of 3

- February 3rd 2009, 09:13 PM #1

- Joined
- Feb 2009
- Posts
- 18

## [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.

- February 3rd 2009, 09:41 PM #2

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,002
- Thanks
- 822

- February 3rd 2009, 09:51 PM #3