What is the answer to 2^{x+1}=7 ?

The answer to 2^{x+1}=7 is x=(ln(7))/(ln(2))-1

Solve for x: 2^{x+1}=7

The solution is x=(ln(7))/(ln(2))-1

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2^{x + 1} = 7

x = \frac{ln ( 7 )}{ln ( 2 )} - 1

Decimal Notation

x = 1.80735 \dots

Solution steps

2^{x + 1} = 7

Apply exponent rules

(x + 1) ln (2) = ln (7)

Solve

(x + 1) ln (2) = ln (7) : x = \frac{ln ( 7 )}{ln ( 2 )} - 1

x = \frac{ln ( 7 )}{ln ( 2 )} - 1

e^{4 x - 5} = 13

a, x^{a + 1} = \frac{1}{2}

3^{∣ x ∣ - 1} = 9

3 \cdot 2^{x} = 48

3^{2 x - 1} = 7

What is the answer to 2^{x+1}=7 ?
The answer to 2^{x+1}=7 is x=(ln(7))/(ln(2))-1
Solve for x: 2^{x+1}=7
The solution is x=(ln(7))/(ln(2))-1