What is the answer to solvefor x,2^x3^y=6 ?

The answer to solvefor x,2^x3^y=6 is x=(ln(2)+ln(3)(-y+1))/(ln(2))

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solvefor x,2^x3^y=6

solve for

x, 2^{x} 3^{y} = 6

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

Solution steps

2^{x} \cdot 3^{y} = 6

Divide both sides by

3^{y}

2^{x} = \frac{6}{3 ^{y}}

Simplify

\frac{6}{3 ^{y}} : 2 \cdot 3^{1 - y}

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

Apply exponent rules

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

Solve

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

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

3 \cdot 5^{2 t - 1} = 75

4 \cdot 2^{- \frac{t}{5}} = 288

77 = 1 1^{x}

y, e^{x + y} = x

14 (8)^{x} = 7

What is the answer to solvefor x,2^x3^y=6 ?
The answer to solvefor x,2^x3^y=6 is x=(ln(2)+ln(3)(-y+1))/(ln(2))