What is the answer to 4^x-2^{x+3}+12=0 ?

The answer to 4^x-2^{x+3}+12=0 is x=(ln(6))/(ln(2)),x=1

Solve for x: 4^x-2^{x+3}+12=0

The solution is x=(ln(6))/(ln(2)),x=1

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4^{x} - 2^{x + 3} + 12 = 0

x = \frac{ln ( 6 )}{ln ( 2 )}, x = 1

Decimal Notation

x = 2.58496 \dots, x = 1

Solution steps

4^{x} - 2^{x + 3} + 12 = 0

Apply exponent rules

(2^{x})^{2} - 2^{x} \cdot 2^{3} + 12 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - u \cdot 2^{3} + 12 = 0

Solve

u^{2} - u \cdot 2^{3} + 12 = 0 : u = 6, u = 2

u = 6, u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 6 : x = \frac{ln ( 6 )}{ln ( 2 )}

Solve

2^{x} = 2 : x = 1

x = \frac{ln ( 6 )}{ln ( 2 )}, x = 1

2^{x + 3} - 2^{x} = 56

a^{x} = 2

4 e^{3 x - 2} = 8

3^{x - 3} = 27

2^{y + 3} + 2^{y} = 18

What is the answer to 4^x-2^{x+3}+12=0 ?
The answer to 4^x-2^{x+3}+12=0 is x=(ln(6))/(ln(2)),x=1
Solve for x: 4^x-2^{x+3}+12=0
The solution is x=(ln(6))/(ln(2)),x=1