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

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

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

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

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3^{3 x + 1} = 2

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

Decimal Notation

x = - 0.12302 \dots

Solution steps

3^{3 x + 1} = 2

Apply exponent rules

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

Solve

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

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

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

4^{x - 2} = 2^{x}

0. 5^{3 - x} = \frac{64}{4 ^{x - 2}}

x = 2^{- x}

9^{x} = 6

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