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Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35

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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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New post 11 Oct 2016, 09:10
1
dvinoth86 wrote:
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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New post 03 Dec 2016, 20:36
thefibonacci wrote:
dvinoth86 wrote:
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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New post 05 Dec 2016, 01:23
AmritaSarkar89 wrote:
thefibonacci wrote:
dvinoth86 wrote:
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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New post 15 Jul 2017, 08:37
Bunuel wrote:
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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New post 15 Jul 2017, 08:56
Rosicky wrote:
Bunuel wrote:
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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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35  [#permalink]

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Re: Is x > 10^10 ? (1) x > 2^34 (2) x = 2^35   [#permalink] 21 Jun 2019, 21:39

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