Notice that we can write 4^(2x) as (4^x)^2
and you equation becomes a quadratic
let 4^x = k
then your equation becomes:
k^2 - k - 20 = 0
(k-5)(k+4) = 0
k = 5 or k = -4
then 4^x = 5
log 4^x = log 5
x log4 = log 5
x = log5/log4 = appr 1.161
and 4^x = -4
x log4 = log (-4), but log -4 would be undefined , so this part leads to an extraneous root.
x = appr 1.161
Solve the equation. Check for extraneous roots.
4^2x-4^x-20=0
1 answer