Question

solve
4^(2x)-20*4^x+64=0

Answer 1

solve
\[u^2-20u+64=0\] and then replace u by
\[4^x\]
if the solution is negative, ignore it because
\[4^x>0\]

Answer 2

wait, what about u?

Answer 3

\[u^2-20u+64=0\]
\[(u-16_)(u-4)=0\]
\[u=16,u=4\]
\[4^x=16\iff x=2\]
\[4^x=2\iff x=\frac{1}{2}\]

Answer 4

isnt the second one 4 not 2?