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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 11 Oct 2016, 09:10
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dvinoth86
Is x > 10^10 ?

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

Statement 1: x > 2^34

Comparing 2^34 with 10^10
Comparing 2^34 with 2^10 * 5^10 [Now we can cancel out common terms 2^10 from both sides]
Comparing 2^24 with 5^10
Comparing (2^2)^12 with 5^10
Comparing (4)^12 with (5)^10
Comparing (4^6)^2 with (5^5)^2 [Now we can cancel out common powers 2 from both sides]
Comparing (4)^6 with (5)^5
Comparing (4096) with (3125)

Since 4096 > 3125
therefore, 2^34 > 10^10
SUFFICIENT

Statement 2: x = 2^35

Since I know the exact value of x so a comparison can be established hence the statement is sufficient
SUFFICIENT

Answer: Option D
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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 03 Dec 2016, 20:36
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dvinoth86
Is x > 10^10 ?

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

whenever there is a 10 (or its multiples) involved, i use log.

is x > 10^10
or log x > 10 log 10
or log x > 10 ?

1) x > 2^34
log x > 34*log 2
log x > 10.23 (sufficient)

2) has to be sufficient

As an Indian I totally understand resorting to logs but Log is not tested in GMAT so are there chances we can come across such obnoxious calculation extensive questions.
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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 05 Dec 2016, 01:23
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dvinoth86
Is x > 10^10 ?

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

whenever there is a 10 (or its multiples) involved, i use log.

is x > 10^10
or log x > 10 log 10
or log x > 10 ?

1) x > 2^34
log x > 34*log 2
log x > 10.23 (sufficient)

2) has to be sufficient

As an Indian I totally understand resorting to logs but Log is not tested in GMAT so are there chances we can come across such obnoxious calculation extensive questions.

This question can be solved without logs as shown in many solutions on previous pages. For example here: is-x-10-10-1-x-2-34-2-x-127881.html#p1047347
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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 15 Jul 2017, 08:37
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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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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 15 Jul 2017, 08:56
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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.
____________________
Yes.

Please check this to format properly: https://gmatclub.com/forum/rules-for-po ... l#p1096628
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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 06 Aug 2020, 02:04
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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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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 28 Oct 2020, 10:41
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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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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 02 Nov 2021, 02:49
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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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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35 27 Dec 2024, 03:54
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