What is the answer to 2^{x-1}=31 ?

The answer to 2^{x-1}=31 is x=(ln(31))/(ln(2))+1

Solve for x: 2^{x-1}=31

The solution is x=(ln(31))/(ln(2))+1

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2^{x - 1} = 31

x = \frac{ln ( 31 )}{ln ( 2 )} + 1

Solution steps

2^{x - 1} = 31

Apply exponent rules

(x - 1) ln (2) = ln (31)

Solve

(x - 1) ln (2) = ln (31) : x = \frac{ln ( 31 )}{ln ( 2 )} + 1

x = \frac{ln ( 31 )}{ln ( 2 )} + 1

e^{8 - 4 x} = 7

2 \cdot 4^{x} + 2^{3 - 2 x} = 17

0 = e^{x} + x e^{x}

3^{x} = 2187

What is the answer to 2^{x-1}=31 ?
The answer to 2^{x-1}=31 is x=(ln(31))/(ln(2))+1
Solve for x: 2^{x-1}=31
The solution is x=(ln(31))/(ln(2))+1