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

Question

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 - \frac{1}{2}

*kudos for all correct solutions

seed · Accepted Answer

A)  y – 2 
B)  \frac{y - 1}{2} 
C)  2y – 2 
D)  \frac{y}{2} – 2 
E)  2y - \frac{1}{2}

2^x = 5 
 4^y = 20  ->  2^2^y = 20  ->  2^2^y = 2^2 * 5  ->  2^2^y = 2^2 * 2^x  ->  2^2^y = 2^2^+^x

2y = 2+ x
x = 2y - 2

Answer: (C).

PP777 · Answer

Given: 2^x = 5 ; 4^y = 20
To find : x in terms of y

Approach: 2^x = 5.......................................1
 4^y = 20 => 2^2y = (2^2) *5.............................2

Put the value of 5 from eqn 1 to eqn 2, Therefore 2^2y = 2^2 * 2^x

or, 2y= 2+x
so, x= 2y-2

Correct Answer= C

pushpitkc · Answer

Given:    2^x = 5  -> (1) |  4^y = 20  -> (2)

Dividing (2) by (1), we get  \frac{2^{2y}}{2^x} = \frac{20}{5} = 4  ->  2^{2y-x} = 2^2  ->  2y-x = 2

Therefore, the value of x in terms of y is  2y - 2(Option C)

generis · Answer

4^{y}=20  (given)

(2^2)^{y}=2^2*5

\frac{2^{2y}}{2^2}=5

2^{(2y-2)} = 5  
 2^x = 5  (given)
LHS expressions both = 5. Set them equal:
 2^{x}=2^{(2y-2)} 
 x = 2y - 2

Answer C

akadiyan · Answer

Known Information: 2x=5 and 4y=20

Dividing the equations , we get

\frac{4y}{2x}  =  \frac{20}{5}

2^(2y-x)  = 4

2^(2y-x)  =  2^2

2y-x = 2

x=2y-2

Ans:   Option C

PKN · Answer

We can write  4^y = 5* 2^2 = 2^x * 2^2 = 2^(x+2)

so  2^(2y) = 2^(x+2) 
implies that x+2=2y

so, x=2y-2.

Option. (C)

Kudos for correct explanation.

BrentGMATPrepNow · Answer

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
 https://www.youtube.com/watch?v=At7a1ibdOTc

Cheers,
Brent

BrentGMATPrepNow · Answer

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

SSM700plus · Answer

This is 1st equation -  2^x = 5

This is 2nd equation -  4^y = 20

Divide the 1st equation by 2nd equation.

\frac{(2^x)}{(4^y)}= \frac{5}{20}

\frac{(2^x)}{(2^2^y)}= \frac{1}{4}

Reverse the order, change the denominator & numerator
 \frac{(2^2^y)}{(2^x)}= 4

\frac{(2^2^y)}{(2^x)}= 2^2

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

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

Equating powers --> 2y = x +2.

x= 2y-2.

Ans - C

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

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