2^x * 3^y = 72, what is x + y?

Solve for x and solve for y

2^x*3^y=72

divide both sides by 3^y

2^x=72 / (3^y)

take the log₂ of both sides

x=log₂ (72 / (3^y))

x=log₂72-log₂3^y

x=log₂ (18*2^2) - ylog₂3

x=log₂18+log₂2^2-ylog₂3

x=log₂18+2log₂2-ylog₂3

x=log₂18+2-ylog₂3

solve for y

2^x*3^y=72

divide both sides by 2^x

3^y=72 / (2^x)

take log₃ of both sides

y=log₃ (72 / (2^x))

y=log₃72-log₃2^x

y=log₃ (3^2*8) - xlog₃2

y=log₃3^2+log₃8-xlog₃2

y=2log₃3+log₃8-xlog₃2

y=2+log₃8-xlog₃2

so x+y=log₂18+2-ylog₂3+2+log₃8-xlog₃2

x+y=log₂18-ylog₂3+log₃8-xlog₃2+4

not sure how to simplify further

Question: 2^x * 3^y = 72, what is x + y? Solution: x=3 and y=2 / 2^x * 3^y = 72 = 2^3 * 3^2 = 72 So, if x=3 and y=2 then, 3+2 = 5 The answer: x+y = 5