What is the answer to solvefor y,(3^{y+1})/(32)=sqrt(96^x) ?

The answer to solvefor y,(3^{y+1})/(32)=sqrt(96^x) is y=(5ln(2)x+10ln(2)+ln(3)x)/(2ln(3))-1

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solvefor y,(3^{y+1})/(32)=sqrt(96^x)

solve for

y, \frac{3 ^{y + 1}}{32} = 9 6^{x}

y = \frac{5 ln ( 2 ) x + 10 ln ( 2 ) + ln ( 3 ) x}{2 ln ( 3 )} - 1

Solution steps

\frac{3 ^{y + 1}}{32} = 9 6^{x}

Multiply both sides by

32

\frac{3 ^{y + 1}}{32} \cdot 32 = 9 6^{x} \cdot 32

Simplify

3^{y + 1} = 9 6^{x} \cdot 32

Apply exponent rules

(y + 1) ln (3) = ln (9 6^{x \frac{1}{2}} \cdot 32)

Solve

(y + 1) ln (3) = ln (9 6^{x \frac{1}{2}} \cdot 32) : y = \frac{5 ln ( 2 ) x + 10 ln ( 2 ) + ln ( 3 ) x}{2 ln ( 3 )} - 1

y = \frac{5 ln ( 2 ) x + 10 ln ( 2 ) + ln ( 3 ) x}{2 ln ( 3 )} - 1

3^{3 x + 2} = 3

y, x = \frac{1 + e ^{y}}{1 - e ^{y}}

5 (2)^{8 t} = 15

9 e^{x} = 1

x, 4^{x + 1} + 4^{x} = 160

What is the answer to solvefor y,(3^{y+1})/(32)=sqrt(96^x) ?
The answer to solvefor y,(3^{y+1})/(32)=sqrt(96^x) is y=(5ln(2)x+10ln(2)+ln(3)x)/(2ln(3))-1