1. 4^x - 3(4^-x) = 8

Question

I got where 4^x = 4 + square root (19), but I can't solve further. The answer is (log (4 + root (19)) ) / (log 4).

2. log (x^2) = (log x)^2
Solve for x

Thanks!

Answer 1

If 4^x = 4+√19, then
take logs, recalling that log(x^n) = n*log(x)
x*log(4) = log(4+√19)
and so on

log(x^2) = (log x)^2
2 logx = (logx)^2
logx(2-logx) = 0
logx = 2 or 0
x = b^2 or 1
to whatever base b you are using for the logs.