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

The answer to 2^{3x+1}=3^{2-x} is x=(2ln(3)-ln(2))/(ln(24))

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

The solution is x=(2ln(3)-ln(2))/(ln(24))

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

x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 24 )}

Decimal

x = 0.47326 \dots

Solution steps

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

Apply exponent rules

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

Solve

(3 x + 1) ln (2) = (2 - x) ln (3) : x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 24 )}

x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 24 )}

\frac{1}{6} = 6^{x}

7^{3 x - 3} = 343

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

e^{x} - e^{- x} = - 1

1 6^{8^{- 2 7^{- x}}} = 4

What is the answer to 2^{3x+1}=3^{2-x} ?
The answer to 2^{3x+1}=3^{2-x} is x=(2ln(3)-ln(2))/(ln(24))
Solve for x: 2^{3x+1}=3^{2-x}
The solution is x=(2ln(3)-ln(2))/(ln(24))