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: .

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

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- Feb 2009
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## [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

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