Question 1004328
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.