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 – Creator of greenlighttestprep.com

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