If x and y are positive, what is x+y? $$(1)\ \ 2^x\cdot3^y=72.$$ $$(2)\ \ \ 2^x\cdot2^y=32.$$
The OA is B .
Why is not sufficient the statement (1)? Can any expert help me?
Thanks in advanced.
If x and y are positive, what is x+y?
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Hello Vjesus12.VJesus12 wrote:If x and y are positive, what is x+y? $$(1)\ \ 2^x\cdot3^y=72.$$ $$(2)\ \ \ 2^x\cdot2^y=32.$$
The OA is B .
Why is not sufficient the statement (1)? Can any expert help me?
Thanks in advanced.
Let's start with the statement (2): $$2^x\cdot2^y=32\ \Rightarrow\ 2^{x+y}=2^5\ \Rightarrow\ \ \ x+y=5.$$ Hence, this is sufficient.
Now, let's use the statement (1):
If we pick x=3 and y=2 then x+y=5.
But, since x and y not necessary are integers, we can set x=1 and then $$2^x\cdot3^y=72\ \Rightarrow\ \ 2\cdot3^y=72\ \Rightarrow\ \ \ 3^y=36\ \Rightarrow\ \ y\ln\left(3\right)=\ln\left(36\right)\ \Rightarrow\ \ y=\frac{\ln\left(36\right)}{\ln\left(3\right)}.$$ This implies that $$x+y=1+\frac{\ln\left(36\right)}{\ln\left(3\right)}\ne5$$ Therefore, this statement is not sufficient.
This is why the correct answer is B.
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We can approach Statement 1 by thinking about the possible values of x and y:
Integers x =3 and y = 2 make this equation work out really nicely and give us x + y = 5:
$$2^3\cdot3^2=72$$ $$8\cdot9=72$$
However, there is nothing in the problem that indicates x and y must be integers - we only know that they need to be positive. So, for example, if x = 2, then we have
$$2^2\cdot3^y=72$$ $$4\cdot3^y=72$$ $$3^y=18$$
Which makes y some decimal between 2 and 3 - no need to apply any logarithms to figure out what it will be. Because x is an integer and y is a decimal, we know that x + y will not equal integer 5 - it will be some decimal between 4 and 5.
This means that there are at least two possible values for x + y: 5 and some decimal between 4 and 5. This is enough to tell us that Statement 1 is insufficient.
Integers x =3 and y = 2 make this equation work out really nicely and give us x + y = 5:
$$2^3\cdot3^2=72$$ $$8\cdot9=72$$
However, there is nothing in the problem that indicates x and y must be integers - we only know that they need to be positive. So, for example, if x = 2, then we have
$$2^2\cdot3^y=72$$ $$4\cdot3^y=72$$ $$3^y=18$$
Which makes y some decimal between 2 and 3 - no need to apply any logarithms to figure out what it will be. Because x is an integer and y is a decimal, we know that x + y will not equal integer 5 - it will be some decimal between 4 and 5.
This means that there are at least two possible values for x + y: 5 and some decimal between 4 and 5. This is enough to tell us that Statement 1 is insufficient.
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