Question 604691
log[4](x)+ log[4](6x-7)=log[4]5
log[4](x)(6x-7)=log[4]5
(x)(6x-7)=5
6x^2 - 7x = 5
6x^2 - 7x - 5 = 0
6x^2 - 10x + 3x - 5 = 0
(6x^2 - 10x) + (3x - 5) = 0
2x(3x - 5) + (3x - 5) = 0
(3x - 5)(2x+1) = 0
x = {5/3, -1/2}
Since you can't take the log of a negative number, you can throw out -1/2 (extraneous) leaving:
x = 5/3