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: .
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?