What is the answer to 4^{y-2}= 1/(3^{-y)} ?

The answer to 4^{y-2}= 1/(3^{-y)} is y=(4ln(2))/(2ln(2)-ln(3))

Solve for y: 4^{y-2}= 1/(3^{-y)}

The solution is y=(4ln(2))/(2ln(2)-ln(3))

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4^{y - 2} = \frac{1}{3 ^{- y}}

y = \frac{4 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

Solution steps

4^{y - 2} = \frac{1}{3 ^{- y}}

Apply exponent rules

(y - 2) ln (4) = y ln (3)

Solve

(y - 2) ln (4) = y ln (3) : y = \frac{4 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

y = \frac{4 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

Verify Solutions:

y = \frac{4 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

True

The solution is

y = \frac{4 ln ( 2 )}{2 ln ( 2 ) - ln ( 3 )}

5 = 2^{x}

4^{2 x - 1} = 8^{x + 3}

3^{x + 2} = 7

3^{x} = (\frac{1}{3})^{x + 1}

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

What is the answer to 4^{y-2}= 1/(3^{-y)} ?
The answer to 4^{y-2}= 1/(3^{-y)} is y=(4ln(2))/(2ln(2)-ln(3))
Solve for y: 4^{y-2}= 1/(3^{-y)}
The solution is y=(4ln(2))/(2ln(2)-ln(3))