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

The answer to 2^{3x-2}=3^{x+1} is x=(ln(12))/(3ln(2)-ln(3))

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

The solution is x=(ln(12))/(3ln(2)-ln(3))

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

x = \frac{ln ( 12 )}{3 ln ( 2 ) - ln ( 3 )}

Decimal Notation

x = 2.53347 \dots

Solution steps

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

Apply exponent rules

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

Solve

(3 x - 2) ln (2) = (x + 1) ln (3) : x = \frac{ln ( 12 )}{3 ln ( 2 ) - ln ( 3 )}

x = \frac{ln ( 12 )}{3 ln ( 2 ) - ln ( 3 )}

3^{x} + 3^{x + 4} + 3^{x - 1} = \frac{247}{27}

3^{5 x} = 2 7^{2 x - 4}

2^{x + 3} = 16

\frac{75}{1 + 15 e ^{- 0.4 x}} = 25

7^{x^{2}} = (\frac{1}{49})^{3 x + 4}

What is the answer to 2^{3x-2}=3^{x+1} ?
The answer to 2^{3x-2}=3^{x+1} is x=(ln(12))/(3ln(2)-ln(3))
Solve for x: 2^{3x-2}=3^{x+1}
The solution is x=(ln(12))/(3ln(2)-ln(3))