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

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

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

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

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

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

Decimal Notation

x = 2.70951 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3^{3 x - 2} = 81

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

3^{x} = 7

4 e^{2 x} - 7 e^{x} = 15

2^{2 x} - 5 \cdot 2^{x} + 6 = 0

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