What is the answer to (4^{x-1})^2-10(4^{x-1})+16=0 ?

The answer to (4^{x-1})^2-10(4^{x-1})+16=0 is x= 5/2 ,x= 3/2

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(4^{x-1})^2-10(4^{x-1})+16=0

(4^{x - 1})^{2} - 10 (4^{x - 1}) + 16 = 0

x = \frac{5}{2}, x = \frac{3}{2}

Decimal Notation

x = 2.5, x = 1.5

Solution steps

(4^{x - 1})^{2} - 10 (4^{x - 1}) + 16 = 0

Rewrite the equation with

4^{x - 1} = u

(u)^{2} - 10 u + 16 = 0

Solve

u^{2} - 10 u + 16 = 0 : u = 8, u = 2

u = 8, u = 2

Substitute back

u = 4^{x - 1},

solve for

x

Solve

4^{x - 1} = 8 : x = \frac{5}{2}

Solve

4^{x - 1} = 2 : x = \frac{3}{2}

x = \frac{5}{2}, x = \frac{3}{2}

n, x^{n} = ln (x + 1)

3^{- 5 x + 2} = (\frac{1}{9})^{x + 2}

2 + x = 2^{x}

8 \cdot 2^{3 x} - 32 = 0

What is the answer to (4^{x-1})^2-10(4^{x-1})+16=0 ?
The answer to (4^{x-1})^2-10(4^{x-1})+16=0 is x= 5/2 ,x= 3/2
Solve for x: (4^{x-1})^2-10(4^{x-1})+16=0
The solution is x= 5/2 ,x= 3/2