What is the answer to 2^{2x-1}+8=5(2^x) ?

The answer to 2^{2x-1}+8=5(2^x) is x=3,x=1

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2^{2x-1}+8=5(2^x)

2^{2 x - 1} + 8 = 5 (2^{x})

x = 3, x = 1

Solution steps

2^{2 x - 1} + 8 = 5 (2^{x})

Apply exponent rules

(2^{x})^{2} \cdot 2^{- 1} + 8 = 5 \cdot 2^{x}

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{- 1} + 8 = 5 u

Solve

u^{2} \cdot 2^{- 1} + 8 = 5 u : u = 8, u = 2

u = 8, u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 8 : x = 3

Solve

2^{x} = 2 : x = 1

x = 3, x = 1

e^{2 x} - 6 e^{x} - 7 = 0

2 5^{3 x + 1} = 81

e^{- 2 \cdot x} = 1

2 = 1 0^{2 a}

6^{x} = e^{x + 1}

What is the answer to 2^{2x-1}+8=5(2^x) ?
The answer to 2^{2x-1}+8=5(2^x) is x=3,x=1
Solve for x: 2^{2x-1}+8=5(2^x)
The solution is x=3,x=1