What is the answer to 3^{x+1}=2 ?

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

Solve for x: 3^{x+1}=2

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

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3^{x + 1} = 2

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

Solution steps

3^{x + 1} = 2

Apply exponent rules

(x + 1) ln (3) = ln (2)

Solve

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

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

e^{(6 x)} - 5 e^{(3 x)} + 4 = 0

3^{2 x - 2} + 3^{x + 1} = 12

2^{2 x - 1} - 5 \cdot 2^{x + 1} + 32 = 0

1^{x} = 0

1^{x} = 1

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