SOLUTION: log7(8x)=log7(2x)+log7(3x-5)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: log7(8x)=log7(2x)+log7(3x-5)      Log On


   

Question 1004328: log7(8x)=log7(2x)+log7(3x-5)
Found 2 solutions by stanbon, fractalier:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log7(8x)=log7(2x)+log7(3x-5)
-----
log7(8x) = log7[2x(3x-5)]
-----
8x = 6x^2-10x
-----
6x^2 - 18x = 0
----
6x(x-3) = 0
----
x = 0 or x = 3
------------------
Acceptable answer:: x = 3
-----------------------
Cheers,
Stan H.

Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
I imagine you wish to solve this...
log7(8x)=log7(2x)+log7(3x-5)
log7(8x)=log7[(2x)(3x-5)]
Now exponentiate 7-to-the
8x = (2x)(3x-5)
8x = 6x^2 - 10x
6x^2 - 18x = 0
x^2 - 3x = 0
x(x-3) = 0
x = 0 or x = 3
You cannot take the log of zero, so x = 3.