What is the answer to 16-5*2^x+4^{x-1}=0 ?

The answer to 16-5*2^x+4^{x-1}=0 is x=4,x=2

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16-5*2^x+4^{x-1}=0

16 - 5 \cdot 2^{x} + 4^{x - 1} = 0

x = 4, x = 2

Solution steps

16 - 5 \cdot 2^{x} + 4^{x - 1} = 0

Apply exponent rules

16 - 5 \cdot 2^{x} + (2^{x})^{2} \cdot 2^{- 2} = 0

Rewrite the equation with

2^{x} = u

16 - 5 u + (u)^{2} \cdot 2^{- 2} = 0

Solve

16 - 5 u + u^{2} \cdot 2^{- 2} = 0 : u = 16, u = 4

u = 16, u = 4

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 16 : x = 4

Solve

2^{x} = 4 : x = 2

x = 4, x = 2

9^{x} = 3^{x - 1}

x, 4^{x} = 10 - 4^{x + 1}

3 e^{3 t} - 2 = 0

3^{x^{2}} = 81

8 = 2^{5 x + 7}

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