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

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

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

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

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

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

Decimal Notation

x = 1.70951 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

2^{x + 1} = 64

e^{x} - 6 e^{- x} = 1

\frac{1}{27} = 3^{n}

e^{2 x} = m - 3

9 e^{x - 2} = 20

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