What is the answer to 3*2^{x+1}-64=0 ?

The answer to 3*2^{x+1}-64=0 is x=(ln(64/3))/(ln(2))-1

Solve for x: 3*2^{x+1}-64=0

The solution is x=(ln(64/3))/(ln(2))-1

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3 \cdot 2^{x + 1} - 64 = 0

x = \frac{ln ( \frac{64}{3} )}{ln ( 2 )} - 1

Decimal Notation

x = 3.41503 \dots

Solution steps

3 \cdot 2^{x + 1} - 64 = 0

Add

64

to both sides

3 \cdot 2^{x + 1} - 64 + 64 = 0 + 64

Simplify

3 \cdot 2^{x + 1} = 64

Divide both sides by

3

2^{x + 1} = \frac{64}{3}

Apply exponent rules

(x + 1) ln (2) = ln (\frac{64}{3})

Solve

(x + 1) ln (2) = ln (\frac{64}{3}) : x = \frac{ln ( \frac{64}{3} )}{ln ( 2 )} - 1

x = \frac{ln ( \frac{64}{3} )}{ln ( 2 )} - 1

3^{x - 2} = 3^{17}

25 (5^{2 x}) + 24 (5^{x}) - 1 = 0

9^{x} = (\frac{1}{3})^{x^{2} - x}

- 1^{x} = 1

3^{x} + 4 \cdot 3^{x + 1} = 13

What is the answer to 3*2^{x+1}-64=0 ?
The answer to 3*2^{x+1}-64=0 is x=(ln(64/3))/(ln(2))-1
Solve for x: 3*2^{x+1}-64=0
The solution is x=(ln(64/3))/(ln(2))-1