What is the answer to solvefor x,3^{3x}=7^{2x+1} ?

The answer to solvefor x,3^{3x}=7^{2x+1} is x=(ln(7))/(3ln(3)-2ln(7))

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solvefor x,3^{3x}=7^{2x+1}

solve for

x, 3^{3 x} = 7^{2 x + 1}

x = \frac{ln ( 7 )}{3 ln ( 3 ) - 2 ln ( 7 )}

Decimal Notation

x = - 3.26504 \dots

Solution steps

3^{3 x} = 7^{2 x + 1}

Apply exponent rules

3 x ln (3) = (2 x + 1) ln (7)

Solve

3 x ln (3) = (2 x + 1) ln (7) : x = \frac{ln ( 7 )}{3 ln ( 3 ) - 2 ln ( 7 )}

x = \frac{ln ( 7 )}{3 ln ( 3 ) - 2 ln ( 7 )}

(- 0.2)^{x} = 38 (\frac{1}{3})^{x}

1 1^{x} = 121

5 x = 2 e^{x + 1}

3^{x - 4} = 9

3^{x - 4} = 5

What is the answer to solvefor x,3^{3x}=7^{2x+1} ?
The answer to solvefor x,3^{3x}=7^{2x+1} is x=(ln(7))/(3ln(3)-2ln(7))