GIVEN: \(4^x=100\)
Notice the following:
\(4^3=64\)
\(4^4=256\)
Since 100 is BETWEEN 64 and 256, it must be the case that x is BETWEEN 3 and 4.
So let's say x = 3.something

GIVEN: \(2^y=50\)
Notice the following:
\(2^5=32\)
\(2^6=64\)
Since 50 is BETWEEN 32 and 64, it must be the case that y is BETWEEN 5 and 6.
So let's say y = 5.something

This means x - y ≈ 3.something - 5.something ≈ -2

Answer: D
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Asked: Based on the two equations below, the value of x - y is closest to which number?

\(4^x=100\)
\(2^y=50\)

x log 4 = log 100
2x log 2 = 2 log 10
x = log 10/log 2 = 1/.3010 = 3.32

y log 2 = log 50 = log 100 - log 2 = 2 - log 2
y = (2-.3010)/.3010 = 5.64

x - y = 3.32 - 5.64 = - 2.32

IMO D
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Based on the two equations below, the value of x - y is closest to which number?

\(4^x=100\)
\(2^y=50\)

\(4^x=50*2\)

\(4^x=2^y*2\)

\(2^(2x)=2^(y+1)\)

2x = y+1
\(x-y =\frac{ (y+1)}{2} - y\)
\(x-y =\frac{ (1-y)}{2}\)


since \(2^y=50\); 5<y<6
-6< -y <-5
-5<1-y<-4
\(-2.5< \frac{(1-y)}{2} <-2\)

Option D is the closest to the range.

Three responses as of the reveal, and three separate ways of tackling the same problem. Thank you, BrentGMATPrepNow, Kinshook, and sumitkrocks. I will confess to designing the problem by following the logic outlined by Brent. Of course, logarithms would provide a method to pin down the exact values of the two unknowns, but I was curious to see what people might come up with in a GMAT™ setting. I never would have thought of the algebraic method in the post just above. In any case, I will avoid treading on the heels of Brent, but I will add a bit in the same vein.

Of course, we can determine with relative ease that x is between 3 and 4, and also that y is between 5 and 6. This information restricts the values of the difference we are seeking to the following:

\(3 < x < 4\)

\(5 < y < 6\)

\(3 - 6 < x - y < 4 - 5\) or

\(-3 < x - y < -1\)

But for the inner fact-checker, how do we feel comfortable getting behind any of the answer choices (C), (D), or (E)? Which one is it? This is where I use estimation and logic to zero in on the target. Consider the ranges between our low values (the outcome if each low value were inserted into the equations), the known values from the equations, and the high values that would be derived:

\(4^x=100\)

\(4^3=64\)

\(4^4=256\)

Graphically, that might resemble the following (with the known value in bold):

64----100----------256

It stands to reason that since 64 is closer to 100 than is 256, the value of x will be closer to 3 than to 4. We can ballpark it anywhere from, say, 3.2 to 3.4. (We do not care about the specifics.) How about 3.3, to be conservative?

Now, repeat the process with the other variable:

\(2^y=50\)

\(2^5=32\)

\(2^6=64\)

32---50--64

Although this one is a little tighter, we can see that 50 is closer to 64 than to 32, so the value of y will be closer to 6 than to 5. We can call it 5.6 if we want.

\(x-y=\)

\(3.3-5.6=-2.3\)

We can feel much better about selecting (D), even if we are not quite sure what, exactly, the values of x and y may be. The point is that you can use the method I have just outlined to approximate the values of exponents and put yourself in a good spot within a short amount of time. Nothing beats knowing you have the answer inside a minute or so.

Happy studies to everyone.

- Andrew

Given:
2^2x ~ 2^6(100)
2x ~ 6
x ~ 3
2^y ~ 2^5(50)
y ~ 5
x - y ~ 3 - 5 ~ -2
IMO, (D)!
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\(4^3 = 64 , 4^4 = 256\) So, \(4^x = 100 => 3.k\)
\(2^5 = 32 , 2^6 = 64\) So, \(2^x = 50 => 5.k\)


So, \(x - y => 3.k - 5.4 = -2.k\), Answer must be (D)
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