Question

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

Answer 1

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

Answer 2

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