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

The answer to 2^{x+1}=3^{x-1} is x=(ln(6))/(ln(3)-ln(2))

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

The solution is x=(ln(6))/(ln(3)-ln(2))

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

x = \frac{ln ( 6 )}{ln ( 3 ) - ln ( 2 )}

Solution steps

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

Apply exponent rules

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

Solve

(x + 1) ln (2) = (x - 1) ln (3) : x = \frac{ln ( 6 )}{ln ( 3 ) - ln ( 2 )}

x = \frac{ln ( 6 )}{ln ( 3 ) - ln ( 2 )}

(\frac{1}{5})^{3 x} = 125

4^{x - 2} = 4^{x} - 4^{2}

5^{x - 2} = 1

2^{x - 1} + 3 = 7

2^{x} - 2 + 2^{- x} = 0

What is the answer to 2^{x+1}=3^{x-1} ?
The answer to 2^{x+1}=3^{x-1} is x=(ln(6))/(ln(3)-ln(2))
Solve for x: 2^{x+1}=3^{x-1}
The solution is x=(ln(6))/(ln(3)-ln(2))