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Bunuel
Is x > 10^10 ?

(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.

OR: \(2^{34}=(2^{10})^{3.4}=(1,024)^{3.4}>(10^3)^{3.4}=10^{(3*3.4)}=10^{10.2}>10^{10}\).

(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.

Answer: D.


Hi Bunuel,

In statement 1, I'm a bit confused. If x >y, is it necessary that under _/x > _/y (<< those are under roots )

Thanks.
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Bunuel
Is x > 10^10 ?

(1) x > 2^34 --> we should compare \(2^{34}\) and \(10^{10}\) --> take the square root from both: we should compare \(2^{17}\) and \(10^5=100,000\). Now, \(2^{17}=2^{10}*2^7=1,024*128>100,000\). Sufficient.

OR: \(2^{34}=(2^{10})^{3.4}=(1,024)^{3.4}>(10^3)^{3.4}=10^{(3*3.4)}=10^{10.2}>10^{10}\).

(2) x = 2^35. Since we have the exact numerical value of x we should be able compare it to 10^10 and answer the question. It really doesn't matter whether 2^35>10^10, the main point is that we have sufficient information to get the answer. Sufficient.

Answer: D.


Hi Bunuel,

In statement 1, I'm a bit confused. If x >y, is it necessary that under _/x > _/y (<< those are under roots )

Thanks.
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Yes.

Please check this to format properly: https://gmatclub.com/forum/rules-for-po ... l#p1096628
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USE LOGARITHMS:

I was solving this question and as soon as I saw the values 10^10 and 2^34 and 2^35, my brain instantly goes to logarithms

TIP:
Use these relations
Log(mn) = log m + log in
Log(m/n)= log m - log m
Log(10^n) = n
Log(m^n) = n log m
AND
remember these values
log 2 = 0.301
log 3 = 0.477
log 5 = 0.699
log 7 = 0.845

Split the number into different combinations and it will help you find log values
(This wont help for higher prime numbers like log13 or log23 etc but such expressions are seldom asked to evaluate)


BACK TO THE QUESTION:
10^10 = 10*log(10) = 10*1 = 10
>(2^34) = >{34*log(2)} = >(34*0.301) = >10.234
2^35 = 35*log(2) = 35*0.301 = 10.535

Hence each statement is sufficient
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\(2^{34} > 10^{10}\)
\(2^{17} > 10^5\) -- take the square root of both sides
\(2^{17} > 2^{5}5^{5}\) -- factor the right side
\(2^{12} > 5^{5}\)
\(4,096 > 3,125\)

Statement 1 is sufficient.

Statement 2 is clearly sufficient.

Answer is D.
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dvinoth86
Is x > 10^10 ?

(1) x > 2^34
(2) x = 2^35

Goal here is to convert (1) and (2) individually closer/similar to 10^10, so that we can compare easily:

(1) x > 2^34
2^34 = (2^10)^3.4 = (1024)^3.4 > ((10)^3)^3.4 or 10^10.2
Hence, x > 2^34 > 10^10.2, which is obviously > 10^10.
Sufficient

(2) No need to check as (1) already showed us that 2^34 > 10^10.
Hence, Sufficient

Answer: D
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