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

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

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

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

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

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

Decimal Notation

x = - 1.65211 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

2 7^{x} = \frac{9}{( 3 ^{x} )}

3^{x} = 6

3^{x} = 5

5^{x} = 4^{x + 1}

9^{x} - 2 \cdot 3^{x + 2} + 81 = 0

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