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# If 2^x = 5 and 4^y = 20, what is the value

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If 2^x = 5 and 4^y = 20, what is the value [#permalink]  01 Jun 2018, 07:03
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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
[Reveal] Spoiler: OA

_________________

Brent Hanneson – Creator of greenlighttestprep.com

Last edited by Carcass on 02 Jun 2018, 01:55, edited 2 times in total.
Edited by Carcass
GRE Instructor
Joined: 10 Apr 2015
Posts: 3908
Followers: 164

Kudos [?]: 4775 [0], given: 70

Re: If 2^x = 5 and 4^y = 20, what is the value [#permalink]  02 Jun 2018, 05:17
Expert's post
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) $$2y + 2$$

There are several approaches we can take. Here's an algebraic approach

Given: 2^x = 5 and 4^y = 20

Take second equation and rewrite 4 as 2² to get: (2²)^y = 20
Simplify to get: 2^(2y) = 20

We now have:
2^(2y) = 20
2^x = 5

This means we can write: 2^(2y)/2^x = 20/5
Simplify: 2^(2y - x) = 4
Rewrite as: 2^(2y - x) = 2^2
So, it must be the case that 2y - x = 2
Add x to both sides: 2y = x + 2
Subtract 2 from both sides: 2y - 2 = x

RELATED VIDEO

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

GRE Instructor
Joined: 10 Apr 2015
Posts: 3908
Followers: 164

Kudos [?]: 4775 [0], given: 70

Re: If 2^x = 5 and 4^y = 20, what is the value [#permalink]  02 Jun 2018, 05:31
Expert's post
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 ≈ 2.3

GIVEN: 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 ≈ 2.1

ASIDE: 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 2.3

A) 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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Re: If 2^x = 5 and 4^y = 20, what is the value [#permalink]  14 Jun 2018, 14:57
Even though an explanation is given, I just wanted to post mine to see if it's also a correct method and maybe someone else had a similar train of thought.

$$2^x = 5$$ and $$4^y = 20$$

$$2^x = 5$$ and $$2^2^y = 2^2 * 5$$

Here we recognize that $$2^x = 5$$ and substitute it into the second equation.

$$2^2^y = 2^2 * 2^x$$

$$2^2^y = 2^x ^+ ^2$$

$$2y = x+2$$

$$2y - 2 = x$$

[Reveal] Spoiler:
Re: If 2^x = 5 and 4^y = 20, what is the value   [#permalink] 14 Jun 2018, 14:57
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