What is the answer to 6(3^{2x})=2^{x+1} ?

The answer to 6(3^{2x})=2^{x+1} is x=-(ln(3))/(2ln(3)-ln(2))

Solve for x: 6(3^{2x})=2^{x+1}

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

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6 (3^{2 x}) = 2^{x + 1}

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

Decimal Notation

x = - 0.73042 \dots

Solution steps

6 (3^{2 x}) = 2^{x + 1}

Apply exponent rules

ln (6) + 2 x ln (3) = (x + 1) ln (2)

Solve

ln (6) + 2 x ln (3) = (x + 1) ln (2) : x = - \frac{ln ( 3 )}{2 ln ( 3 ) - ln ( 2 )}

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

9^{2 x - 1} = 27

ln (3) = e^{x}

e^{x} - 6 e^{- x} = 0

(2^{x} - 3) (2^{x} - 4) = 0

9^{- x + 15} = 2 7^{x}

What is the answer to 6(3^{2x})=2^{x+1} ?
The answer to 6(3^{2x})=2^{x+1} is x=-(ln(3))/(2ln(3)-ln(2))
Solve for x: 6(3^{2x})=2^{x+1}
The solution is x=-(ln(3))/(2ln(3)-ln(2))