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

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

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

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

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

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

Decimal Notation

x = - 0.73042 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

3^{x - 1} = \frac{3}{27}

e^{3 a} = 8

8^{x - 3} = 4^{3 x - 1}

e^{3 x - 2} = 1

1000 = 50 \cdot 3^{\frac{x}{2}}

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