PrecalculusChallengeExponential and Logarithmic Functions ShareExponential and Logarithmic Functions · Challenge problemSolve the base-tower equation 2 2x−5⋅2 2 x−1+4=0\displaystyle 2^{\,2^{x}}-5\cdot 2^{\,2^{\,x-1}}+4=022x−5⋅22x−1+4=0. (Hint: let y=2 2x−1y=2^{\,2^{x-1}}y=22x−1 and note 2 2x=y22^{\,2^{x}}=y^{2}22x=y2.)Check Show answer Walk through it step by step Practice more like this Review this concept