What is the answer to 4^x-5(2^{x+2})+96=0 ?

The answer to 4^x-5(2^{x+2})+96=0 is x=(ln(12))/(ln(2)),x=3

Solve for x: 4^x-5(2^{x+2})+96=0

The solution is x=(ln(12))/(ln(2)),x=3

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4^{x} - 5 (2^{x + 2}) + 96 = 0

x = \frac{ln ( 12 )}{ln ( 2 )}, x = 3

Solution steps

4^{x} - 5 (2^{x + 2}) + 96 = 0

Apply exponent rules

(2^{x})^{2} - 5 \cdot 2^{x} \cdot 2^{2} + 96 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - 5 u \cdot 2^{2} + 96 = 0

Solve

u^{2} - 5 u \cdot 2^{2} + 96 = 0 : u = 12, u = 8

u = 12, u = 8

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 12 : x = \frac{ln ( 12 )}{ln ( 2 )}

Solve

2^{x} = 8 : x = 3

x = \frac{ln ( 12 )}{ln ( 2 )}, x = 3

3^{(1 - 2 x)} = 0

2 5^{x} - 3 \cdot 5^{x} = 10

3^{12 x + 12} = 8 1^{3 x + 3}

x, 4^{5 x + 4} = 3

2^{2 x} - 5 (2^{x}) = - 4

What is the answer to 4^x-5(2^{x+2})+96=0 ?
The answer to 4^x-5(2^{x+2})+96=0 is x=(ln(12))/(ln(2)),x=3
Solve for x: 4^x-5(2^{x+2})+96=0
The solution is x=(ln(12))/(ln(2)),x=3