What is the answer to 4^{x-1}-2^{x+3}+28=0 ?

The answer to 4^{x-1}-2^{x+3}+28=0 is x=(ln(28))/(ln(2)),x=2

Solve for x: 4^{x-1}-2^{x+3}+28=0

The solution is x=(ln(28))/(ln(2)),x=2

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4^{x - 1} - 2^{x + 3} + 28 = 0

x = \frac{ln ( 28 )}{ln ( 2 )}, x = 2

Decimal Notation

x = 4.80735 \dots, x = 2

Solution steps

4^{x - 1} - 2^{x + 3} + 28 = 0

Apply exponent rules

(2^{x})^{2} \cdot 2^{- 2} - 2^{x} \cdot 2^{3} + 28 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} \cdot 2^{- 2} - u \cdot 2^{3} + 28 = 0

Solve

u^{2} \cdot 2^{- 2} - u \cdot 2^{3} + 28 = 0 : u = 28, u = 4

u = 28, u = 4

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 28 : x = \frac{ln ( 28 )}{ln ( 2 )}

Solve

2^{x} = 4 : x = 2

x = \frac{ln ( 28 )}{ln ( 2 )}, x = 2

6^{x - 10} = 4

e^{2 x + 3} = 7

2 x = 2^{x}

3^{- x + 1} = 9^{x + 2}

4^{x} = 17

What is the answer to 4^{x-1}-2^{x+3}+28=0 ?
The answer to 4^{x-1}-2^{x+3}+28=0 is x=(ln(28))/(ln(2)),x=2
Solve for x: 4^{x-1}-2^{x+3}+28=0
The solution is x=(ln(28))/(ln(2)),x=2