GreenlightTestPrep wrote:
If \(2^x = 5\) and \(4^y = 20\), what is the value of x in terms of y?
A) \(y - 2\)
B) \(\frac{(y + 1)}{2}\)
C) \(2y - 2\)
D) \(\frac{y}{2} - 2\)
E) \(2y + 2\)
*kudos for all correct solutions
Here's a totally different approach:
GIVEN: 2^x = 5
2^2 = 4 and 2^3 = 8
Since 5 is BETWEEN 4 and 8, we know that
x is BETWEEN 2 and 3. From here, we can
estimate.
Since 5 is closer to 4 than it is to 8, we know that x will be closer to 2 than it is to 3.
Let's say that x ≈
2.3GIVEN: 4^y = 20
4^2 = 16 and 4^3 = 64
Since 20 is BETWEEN 16 and 64, we know that
y is BETWEEN 2 and 3.
From here, we can
estimate.
Since 20 is closer to 16 than it is to 64, we know that y will be closer to 2 than it is to 3.
Let's say that y ≈
2.1ASIDE: As we'll see, it doesn't matter if our estimates are a little off
Now that we know that x ≈
2.3 and y ≈
2.1, we check the answers to see which one works.
That is, when we replace y with
2.1, which one yields an x-value that's close to
2.3A) y - 2 =
2.1 - 2 =
0.1 This suggests that, when y =
2.1, x =
0.1. We want x =
2.3. ELIMINATE A.
B) y - 1/2 =
2.1 - 0.5 =
1.6 This suggests that, when y =
2.1, x =
1.6. We want x =
2.3. ELIMINATE B.
C) 2y - 2 = 2(
2.1) - 2 = 4.2 - 2.1 =
2.1 This is VERY CLOSE to
2.3. KEEP C.
D) y/2 - 2 =
2.1/2 - 2 =
some negative number We want x =
2.3. ELIMINATE D.
E) 2y - 1/2 = 2(
2.1) - 0.5 =
3.7 This suggests that, when y =
2.1, x =
3.7. We want x =
2.3. ELIMINATE E
By the process of elimination, the correct answer must be C
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep