What is the answer to 2^{x-1}=35 ?

The answer to 2^{x-1}=35 is x=(ln(35))/(ln(2))+1

Solve for x: 2^{x-1}=35

The solution is x=(ln(35))/(ln(2))+1

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2^{x - 1} = 35

x = \frac{ln ( 35 )}{ln ( 2 )} + 1

Decimal Notation

x = 6.12928 \dots

Solution steps

2^{x - 1} = 35

Apply exponent rules

(x - 1) ln (2) = ln (35)

Solve

(x - 1) ln (2) = ln (35) : x = \frac{ln ( 35 )}{ln ( 2 )} + 1

x = \frac{ln ( 35 )}{ln ( 2 )} + 1

a, x^{a} = x^{11} + x^{10}

12 = \frac{100}{2 ^{\frac{x}{10}}}

2^{x} = x

x, 2 a^{x + 1} + 4^{y + 1} = 96

2^{y} = 1.59

What is the answer to 2^{x-1}=35 ?
The answer to 2^{x-1}=35 is x=(ln(35))/(ln(2))+1
Solve for x: 2^{x-1}=35
The solution is x=(ln(35))/(ln(2))+1