What is the answer to 3^{2x+1}=2^{x-4} ?

The answer to 3^{2x+1}=2^{x-4} is x=-(ln(48))/(2ln(3)-ln(2))

Solve for x: 3^{2x+1}=2^{x-4}

The solution is x=-(ln(48))/(2ln(3)-ln(2))

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3^{2 x + 1} = 2^{x - 4}

x = - \frac{ln ( 48 )}{2 ln ( 3 ) - ln ( 2 )}

Decimal Notation

x = - 2.57380 \dots

Solution steps

3^{2 x + 1} = 2^{x - 4}

Apply exponent rules

(2 x + 1) ln (3) = (x - 4) ln (2)

Solve

(2 x + 1) ln (3) = (x - 4) ln (2) : x = - \frac{ln ( 48 )}{2 ln ( 3 ) - ln ( 2 )}

x = - \frac{ln ( 48 )}{2 ln ( 3 ) - ln ( 2 )}

x, e^{x} = - 1

4^{x} - 20 \cdot 2^{x} + 64 = 0

7^{x} = e^{x + 6}

7^{x} = e^{x + 4}

e^{2 x} = \frac{1}{e}

What is the answer to 3^{2x+1}=2^{x-4} ?
The answer to 3^{2x+1}=2^{x-4} is x=-(ln(48))/(2ln(3)-ln(2))
Solve for x: 3^{2x+1}=2^{x-4}
The solution is x=-(ln(48))/(2ln(3)-ln(2))