Precalculus
posted by Lea .
1.Solve 5^(2x) 4(5^x)=12
2. Solve Log[8](log[8]x)=0

let 5^x = y , then we have
y^2  4y  12 = 0
(y6)(x+2) = 0
y = 6 or y = 2
5^x = 6
xlog5 = log6
x = log6/log5 = appr. 1.1133
or
5^x = 2
xlog5 = log (2), but we cannot take logs of negatives, so no solution for that case
x = log6/log5 or 1.1133