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

The answer to 2^{3x+1}=3^{x-2} is x=-(ln(18))/(3ln(2)-ln(3))

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

The solution is x=-(ln(18))/(3ln(2)-ln(3))

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

x = - \frac{ln ( 18 )}{3 ln ( 2 ) - ln ( 3 )}

Decimal Notation

x = - 2.94686 \dots

Solution steps

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

Apply exponent rules

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

Solve

(3 x + 1) ln (2) = (x - 2) ln (3) : x = - \frac{ln ( 18 )}{3 ln ( 2 ) - ln ( 3 )}

x = - \frac{ln ( 18 )}{3 ln ( 2 ) - ln ( 3 )}

5^{x} = 8

4^{x} + 1 6^{x} = 272

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

4 9^{x + 1} + 7^{2 x} = 350

x, 2^{x} = 3^{y}

What is the answer to 2^{3x+1}=3^{x-2} ?
The answer to 2^{3x+1}=3^{x-2} is x=-(ln(18))/(3ln(2)-ln(3))
Solve for x: 2^{3x+1}=3^{x-2}
The solution is x=-(ln(18))/(3ln(2)-ln(3))