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For integers x and y, 2^x + 2^y =2^30. What is the value of x + y?

1/2 * (2^30 + 2^30) = 2^30

2^29 + 2^29 = 2^30

The above equation is similar to 2^x + 2^y =2^30.
so x =29 and y=29

Ans:D
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Bunuel
For integers x and y, 2^x + 2^y =2^30. What is the value of x + y?

A. 30
B. 32
C. 46
D. 58
E. 64

Ans: D

Solution: 2^x + 2^y =2^30
there must be some value of as we know 2^30 only has 2 as its factor so if i take anything common from LHS of the equation the remaining part must be in form of 2 only.
I can take common if and only if x=y; otherwise the remaining part inside the bracket will become odd
for example lets say x= 14 and y = 16 then 2^14+2^16 = 2^14 (1+2^2)= 2^14 * 5 .. and this will happen for all the values of x and y where x is not equal to y
so now=>
the equation becomes 2^x + 2^x =2^30 because x=y
2^x (1+1)= 2^30
2^x * 2 = 2^30
x+1 = 30
x= 29
and y = 29
x+y = 58
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Bunuel
For integers x and y, 2^x + 2^y =2^30. What is the value of x + y?

A. 30
B. 32
C. 46
D. 58
E. 64

Kudos for a correct solution.

IMO: D

Both sides consists of only powers of 2.

Thus in order to that happen x = y must be the case

if x=y then

\(2^x +2^y =2 ^30\) ==> \(2^x +2^x =2 ^30\)
\(2(2^x) =2 ^30\)
x+1 =30
x=29
Thus x+y = 29+29 =58
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2^1 = 2^0 + 2^0
2^2 = 2^1 + 2^1
2^3 = 2^2 + 2^2
.
.
.
2^30 = 2^29 + 2^29

x + y = 58

Option D
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Bunuel
For integers x and y, 2^x + 2^y =2^30. What is the value of x + y?

A. 30
B. 32
C. 46
D. 58
E. 64

Kudos for a correct solution.

2^x+2^y=2^30
Now, 2^30=2^29+2^29=2*2^29
or x+y=29+29=58
Answer D
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Bunuel
For integers x and y, 2^x + 2^y =2^30. What is the value of x + y?

A. 30
B. 32
C. 46
D. 58
E. 64

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

This problem is a good candidate for testing small numbers to find patterns. While your instincts may tell you to simply set x + y equal to 30, small numbers will show that you cannot simply set them equal. If you try \(2^x+2^y=2^6\), for example you can't find values for x + y = 6 that will set the sum equal to 64. The powers of 2 are:

2

4

8

16

32

64

The only pairing that will work is 32 + 32 = 64, meaning that you need \(2^5+2^5=2^6\), which should make sense: 2 times 2^5 will equal 2^6. So this example should teach you that you need to add two \(2^{29}\)s together to get to \(2^{30}\). So x and y are each 29, making the sum x + y equal to 58, answer choice D.
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For integers x and y, 2^x+2^y=2^30. What is the value of x+y?

A. 30
B. 32
C. 46
D. 58
E. 64

You are given 2^x+2^y=2^30 ---> in order to understand the question, try to see what values can y take if you start with fixed values for x ---> remember that x and y MUST be integers.

Thus, lets have x =1 ---> \(2^y=2^{30}-2\) ---> y can not be an integer. Try a bigger value for x.

x = 15 ---> \(2^y=2^{30}-2^{15}\) ---> \(2^y=2^{15}(2^{15}-1)\) , keep trying and you will see that there will always be a 1 inside the bracket except when you write the given expression as

\(2^x+2^y=2^{30}\) --->\(2^x+2^y=2*2^{29}\) ---> \(2^x+2^y=2^{29}+2^{29}\) ---> compare both sides of the equation to get x=y=29 and x+y = 58.

D is thus the correct answer.

Hope this helps.
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Here's a better way to solve the problem (suggested by Anthony Ritz 8-) - big thank you! ).

Step 1: factor out the given equation and see if there is a pattern
\(2^x(1 + 2^{y-x}) = 2^{30}\)
Hm, interesting. We know that \(2^x\) is always even, so \((1 + 2^{y-x})\) should also be even since \(2^{30}\) is even.

For \((1 + 2^{y-x})\) to be an even integer, \(2^{y-x}\) needs to be odd. What situation will \(2^{y-x}\) be an odd number? When \(2^{y-x} = 1\)!
Hence,\(y-x = 0\) and consequently \(x=y\).

Step 2: plug in our findings to the equation
\(2^x + 2^y = 2^x + 2^x = 2 * 2^x = 2^{1+x} = 2^{30}\)
The last two parts of our equation tells us that \(1+x = 30\).
In conclusion, \(x = y = 29\).

Answer: \(x + y = 58\).
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It's actually really easy to spot a pattern here.

2^2 + 2^2 = 8
2^3 = 8
2^4 + 2^4= 16 + 16 = 32
2^5 = 32

So 2^x +2^y = 2^n
to get x+y you just do 2(n-1).
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Bunuel
For integers x and y, \(2^x + 2^y=2^{30}\). What is the value of \(x + y\)?

A. 30
B. 32
C. 46
D. 58
E. 64

Dividing both sides by 2^x, we have:

1 + 2^(y - x) = 2^(30 - x)

We see that the right hand side is an integer power of 2. But the left hand side is 1 more than an integer power of 2, which can happen only if 2^(y - x) is also equal to 1. (When that happens, then the left side becomes 1 + 1 = 2, which is an integer power of 2.) (Also note that the only way in which 2^(y - x) can equal 1 is if (x - y) = 0.) Thus, the two sides of the equation can’t be equal unless y - x = 0, so that the left hand side is also an integer power of 2. So we have

1 + 2^0 = 2^(30 - x)

1 + 1 = 2^(30 - x)

2^1 = 2^(30 - x)

1 = 30 - x

x = 29

Since y - x = 0, so y = 29. Thus x + y = 29 + 29 = 58.

Answer: D
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