What is the value of xy?

Question

(1)  3^x*5^y=75

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

M27-22

Bunuel · Answer

What is the value of xy ? Notice that we are not told that the x and y are integers. (1) 3^x*5^y=75 --> if x and y are integers then as 75=3^1*5^2 then x=1 and y=2 BUT if they are not, then for any value of x there will exist some non-integer y to satisfy given expression and vise-versa (for example if y=1 then 3^x*5^y=3^x*5=75 --> 3^x=25 --> x=some \ irrational \ #\approx{2.9} ). Not sufficient. (2) 3^{(x-1)(y-2)}=1 --> (x-1)(y-2)=0 --> either x=1 and y is ANY number (including 2) or y=2 and x is ANY number (including 1). Not sufficient. (1)+(2) If from (2) x=1 then from (1) 3^x*5^y=3*5^y=75 --> y=2 and if from (2) y=2 then from (1) 3^x*5^y=3^x*25=75 --> x=1 . Thus x=1 and y=2 . Sufficient. Answer: C. P.S. Please read carefully and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention to rule 3. Thank youl.

CrackverbalGMAT · Answer

Since we do not know that x and y are integers hence statement I and II will be insufficient as we can have multiple answers. Combining will be sufficient here as we will have two equations and two answers.

ydmuley · Answer

Hello  Bunuel  - I have a question on this solution, I remember reading somewhere on ur post that GMAT always prefers rational numbers, in that case will this solution still exist?

Not sure if I am getting it wrong?

Posted from my mobile device

Bunuel · Answer

It's not a matter of preferences, If a solution exits, it exists and if it does not then well it does not. Here, for (1) x and y could be irrational, so the statement is not sufficient.

ydmuley · Answer

Hello  Bunuel  - I have a question on this solution, I remember reading somewhere on ur post that GMAT always prefers rational numbers, in that case will this solution still exist?

Not sure if I am getting it wrong?

Posted from my mobile device

It's not a matter of preferences, If a solution exits, it exists and if it does not then well it does not. Here, for (1) x and y could be irrational, so the statement is not sufficient.

Got it thanks Bunuel

Posted from my mobile device

Mo2men · Answer

Dear  Bunuel

I know that 3^0 = 1. But why did you limited  (x-1)(y-2)=0 . Why did not you consider  (x-1)(y-2)=10  or any value. 1^ (any value) = 1.

Thanks

ydmuley · Answer

Hello, let me try to explain here:

Given is -  3^{(x-1)(y-2)}=1

This is only possible when the power of 3 is ZERO.

Hence, we have equate the power of three, in this case

(x-1)(y-2)  as ZERO and it cannot be equal to any other number.

Hope this is clear.

Posted from my mobile device

HisHo · Answer

Bunuel   ydmuley 
Sorry guys, I still have a problem with this:
In statement 2, why we assumed that (x-1)(y-2) is an integer (zero)? it might be any ir/rational number that when powered to 3 equals 1.

Bunuel · Answer

No. Only 3 in the power of 0 equals to 1.

laborumpossimus · Answer

Hi can someone explain why is the answer not D as we get x 1 and y 2 from both the equations how does combining them help us ?

Bunuel · Answer

Your doubt is addressed above. Here it is again:

The issue is that neither statement alone gives both values.

From (1), x = 1 and y = 2 only if x and y are  integers . Since that is not given, many other values are possible.

From (2), you get x = 1 OR y = 2, not both.

Only when you combine the two statements does x = 1 force y = 2, or y = 2 force x = 1.

So the answer is C.

What is the value of xy?

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