What is the answer to 3^{2y+1}=7^y ?

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

Solve for y: 3^{2y+1}=7^y

The solution is y=-(ln(3))/(2ln(3)-ln(7))

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

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

Decimal Notation

y = - 4.37146 \dots

Solution steps

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

Apply exponent rules

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

Solve

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

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

\frac{( 2 ^{x} + 1 )}{( 2 ^{x} - 1 )} = 5

\frac{( 7 ^{3 - 2 x} )}{( 7 ^{7 - 5 x} )} = 4 9^{3 x - 14}

(3^{4})^{x} = 5

1 = 3^{2 x}

\frac{e ^{- t \cdot 1000}}{41.3} = 1

What is the answer to 3^{2y+1}=7^y ?
The answer to 3^{2y+1}=7^y is y=-(ln(3))/(2ln(3)-ln(7))
Solve for y: 3^{2y+1}=7^y
The solution is y=-(ln(3))/(2ln(3)-ln(7))