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

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

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

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

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

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

Decimal Notation

x = - 4.41902 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

4^{m + 1} = \frac{1}{8 2}

6 4^{x + 1} = 4^{x + 7}

A = 50 e^{0.07 \cdot t}

x, e^{x} = 2 e^{- x} + 1

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