What is the answer to solvefor x,2^{x+1}*3^{y-1}=(2*3)^5 ?

The answer to solvefor x,2^{x+1}*3^{y-1}=(2*3)^5 is x=(5ln(2)+ln(3)(-y+6))/(ln(2))-1

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solvefor x,2^{x+1}3^{y-1}=(23)^5

solve for

x, 2^{x + 1} \cdot 3^{y - 1} = (2 \cdot 3)^{5}

x = \frac{5 ln ( 2 ) + ln ( 3 ) ( - y + 6 )}{ln ( 2 )} - 1

Solution steps

2^{x + 1} \cdot 3^{y - 1} = (2 \cdot 3)^{5}

Divide both sides by

3^{y - 1}

2^{x + 1} = \frac{( 2 \cdot 3 ) ^{5}}{3 ^{y - 1}}

Simplify

\frac{( 2 \cdot 3 ) ^{5}}{3 ^{y - 1}} : 32 \cdot 3^{- y + 6}

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

Apply exponent rules

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

Solve

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

x = \frac{5 ln ( 2 ) + ln ( 3 ) ( - y + 6 )}{ln ( 2 )} - 1

\frac{( \frac{( 2 ^{x} )}{16} ) ^{x} - 1}{16} = 0

3^{x - 2} + 3^{x} = 10

x, q = e^{100 x} \cdot v (x)

1.1111 = 1.01 5^{n}

n, s = r \frac{( ( 1 + i ) ^{n} - 1 )}{i}

What is the answer to solvefor x,2^{x+1}*3^{y-1}=(2*3)^5 ?
The answer to solvefor x,2^{x+1}*3^{y-1}=(2*3)^5 is x=(5ln(2)+ln(3)(-y+6))/(ln(2))-1