Question

Solve 4^(x^(2)-1) =8^-x logarithms or your calculator. The larger of the two solutions that you get is

Answer 1

without using logarithms or your calculator

Answer 2

Notice that \(4=2^2\) and \(8=2^3\)
So the equation can be rewritten as
\(2^{(2(x^2-1))}=2^{(2(-x))}\)
\(2^{2x^2-2}=2^{-2x}\)
Since the two have the same base, the exponents must be equal
\(2x^2-2=-2x\)
\(2x^2+2x-2=0\)
\(2(x^2+x-1)=0\)
Now solve \(x^2+x-1=0\) by using quadratic formula :).