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

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

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

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

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

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

Decimal Notation

x = - 1.82678 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

6^{2 x + 3} = 1 5^{x}

7^{(x - 2)} = 5^{x}

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

1 3^{x - 3} = 9

(0.2)^{x} = 25

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