What is the answer to solvefor y,(9^{x-1})/(3^{y+2)}=81 ?

The answer to solvefor y,(9^{x-1})/(3^{y+2)}=81 is y=2x-8

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solvefor y,(9^{x-1})/(3^{y+2)}=81

solve for

y, \frac{9 ^{x - 1}}{3 ^{y + 2}} = 81

y = 2 x - 8

Solution steps

\frac{9 ^{x - 1}}{3 ^{y + 2}} = 81

Apply exponent rules

(- y - 2) ln (3) = ln (\frac{81}{3 ^{2 (x - 1)}})

Solve

(- y - 2) ln (3) = ln (\frac{81}{3 ^{2 (x - 1)}}) : y = 2 x - 8

y = 2 x - 8

Verify Solutions:

y = 2 x - 8

True

The solution is

y = 2 x - 8

3^{4 x + 1} = 8^{x - 5}

5^{2 x} + 5^{x} - 6 = 0

2 5^{x - 3} = 125

4^{x} - 2^{x + 2} = 12

9^{x} = 3

What is the answer to solvefor y,(9^{x-1})/(3^{y+2)}=81 ?
The answer to solvefor y,(9^{x-1})/(3^{y+2)}=81 is y=2x-8