What is the answer to 2^{x-1}= 17/2 ?

The answer to 2^{x-1}= 17/2 is x=(ln(17/2))/(ln(2))+1

Solve for x: 2^{x-1}= 17/2

The solution is x=(ln(17/2))/(ln(2))+1

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2^{x - 1} = \frac{17}{2}

x = \frac{ln ( \frac{17}{2} )}{ln ( 2 )} + 1

Decimal Notation

x = 4.08746 \dots

Solution steps

2^{x - 1} = \frac{17}{2}

Apply exponent rules

(x - 1) ln (2) = ln (\frac{17}{2})

Solve

(x - 1) ln (2) = ln (\frac{17}{2}) : x = \frac{ln ( \frac{17}{2} )}{ln ( 2 )} + 1

x = \frac{ln ( \frac{17}{2} )}{ln ( 2 )} + 1

4 (3 e^{- x} x^{2} - e^{- x} x^{3}) = 0

8 1^{x - 2} = 2 7^{2 x}

3 e^{3 x} - e^{- x} = 0

2^{x + 1} = 1

4^{x} = 3^{\frac{x}{2}}

What is the answer to 2^{x-1}= 17/2 ?
The answer to 2^{x-1}= 17/2 is x=(ln(17/2))/(ln(2))+1
Solve for x: 2^{x-1}= 17/2
The solution is x=(ln(17/2))/(ln(2))+1