Question

If 2^x = 3^y = 6 then which of the following statements is true?

1) x - y = xy 
2) y - x = xy
3) x + y = xy 
4) xy = x/y

Answer 1

Given 2^x = 3^y =6
Taking log on all the sides we get

Log 2^x = Log 3^y = Log6
xLog2 = yLog3 = Log 6
x= Log6/Log2
y= Log6/Log3

We know that Log(ab) = Log a + Log b
therefore Log6 = Log 2 + Log 3

x+y = (Log 6/ Log 2) + (Log 6/Log 3) = Log 6*((Log 2 + Log 3)/Log 2 Log 3) = (Log 6*Log 6/ Log 2* Log 3) = xy

x+y = xy

Answer 2

The answer is x + Y = xy

Answer 3

we have
2^x = 3^y = 6
2^x = 6 and 3^y = 6
multiplying these two,
2^x * 3^y = 36
2^x * 3^y = 4 * 9
2^x =*3^y = 2^2 * 3^2

so,
x=y=2
hence statement (3) is true.

Answer 4

Regardless of the equation, the "Rule of Thumb" in algebra is that you always add (rather than multiply) exponents such that xy = x+y.

Thus the answer is x+y = xy, or #3 is true.