MathCastle
PrecalculusChallengeExponential and Logarithmic Functions

Exponential and Logarithmic Functions · Challenge problem

Solve the base-tower equation 2 2x−5⋅2 2 x−1+4=0\displaystyle 2^{\,2^{x}}-5\cdot 2^{\,2^{\,x-1}}+4=0. (Hint: let y=2 2x−1y=2^{\,2^{x-1}} and note 2 2x=y22^{\,2^{x}}=y^{2}.)