How do I solve 3log10of(x-6)=11 and e^(2x)+10e^(x)-25=0?

Answer 1

3log10of(x-6)=11
log 10 (x-6)^3 = 11
(x-6)^3 = 10^11
x-6 = 10^(11/3)
x = 10^(11/3) + 6

for the second, let e^x = a
then we have
a^2 + 10a - 25 = 0
(a-5)^2 = 0
a = +- 5
e^x = 5 or e^x = -5 (the second has no solution)
x = ln5