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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If 2^x = 5 and 4^y = 20, what is the value
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Updated on: 02 Jun 2018, 01:55
Updated on: 02 Jun 2018, 01:55
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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
A) \(y - 2\)
B) \(\frac{(y + 1)}{2}\)
C) \(2y - 2\)
D) \(\frac{y}{2} - 2\)
E) \(2y + 2\)
*kudos for all correct solutions
Originally posted by GreenlightTestPrep on 01 Jun 2018, 07:03.
Last edited by Carcass on 02 Jun 2018, 01:55, edited 2 times in total.
Last edited by Carcass on 02 Jun 2018, 01:55, edited 2 times in total.
Edited by Carcass
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If 2^x = 5 and 4^y = 20, what is the value
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02 Jun 2018, 05:17
02 Jun 2018, 05:17
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) \(2y + 2\)
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
Answer: C
RELATED VIDEO
Cheers,
Brent
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If 2^x = 5 and 4^y = 20, what is the value
[#permalink]
02 Jun 2018, 05:31
02 Jun 2018, 05:31
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
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
