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Inequality 3

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by DanaJ » Tue Feb 03, 2009 3:54 am
First of all, let's see what 10^10 really means. Since 10 = 2*5, 10^10 = (2*5)^10 = (2^10)*(5^10).
IMHO, this problem requires solid knowledge about powers.

1. You should notice that 128 is greater than 125. 128 = 2^7 and 125 = 5^3. This means that 2^7 > 5^3. Raise this to the 3rd power and you get that 2^21 > 5^9. Now, since 8 > 5 or 2^3 > 5^1, we can use there last two inequalities to solve the problem. You get that:
2^21 > 5^9
2^3 > 5^1
Multiply these two and you get that 2^24 > 5^10.
Use the fact that 2^10 = 2^10 and multiply this with the previous equation to get that 2^(24+10) > (5^10)*(2^10). This is the equivalent of 2^34 > 10^10. Since x > 2^34, then in the end x > 2^34 > 10^10. So 1 is sufficient.

2. You basically use the same line of thought as above to prove that 2^34 > 10^10. Since x = 2^35 is greater than 2^34 (two times greater, to be more specific), then x is certainly greater than 10^10.
So 2 is also sufficient.

Answer D.
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by willbeatthegmat » Tue Feb 03, 2009 6:56 am
danaj u r doin a great job.....is it the only approach to crack this kind of questions?
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by DanaJ » Tue Feb 03, 2009 7:02 am
I really don't know if there are any other approaches... I know this might seem like a very unlikely explanation, but since powers are my thing (i know powers of 2 up to ^13, powers of 3 up to ^6, powers of 5 up to ^4 and some other useful powers - i've learned them by heart because I've had a lot of practice over the years), it's easy to figure out like this... I guess you'll just have to see if anyone else sees any other explanation...
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by Stuart@KaplanGMAT » Tue Feb 03, 2009 10:45 am
willbeatthegmat wrote:danaj u r doin a great job.....is it the only approach to crack this kind of questions?
Dana's approach to statement 1 is great, but we need to do absolutely no work to know that (2) is sufficient.

The question asks if x is bigger than a fixed number.

(2) gives us the exact value for x. Since we know the exact value for x, we can certainly compare it to another fixed number to determine which one is bigger.

Remember, in DS we don't care WHAT the answer is, we just care if we can GET an answer. Therefore, we can say (2) is sufficient even if we don't whether or not it gives us a "yes" or a "no" answer.
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by peddisetty » Tue Feb 03, 2009 5:25 pm
Stuart : You are exactly right. This saves out butt in GMAT. We dont need to solve (2). Great Job. Back to basics.....
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by cjb » Wed Feb 04, 2009 12:25 pm
2^34 = (2^10)^3 * 2^4

= 1024^3 * 16

And that expression must be greater than:

(10^3)^3 * 10
10^9 * 10
10^10

A Maths teacher of mine referred to 2^10 as "the binary thousand".

Once I knew that, it was clear that 2^35 must also be greater, although Stuart's point still applies.
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by GID09 » Fri Feb 06, 2009 10:32 am
Another approach....

10^10 = (2x2x2.5)^10 => 2^20x2.5^10 ...(1)

Case 1 X>2^34 ..so X>2^20 x2^14 as X^a+b =X^a x X^b

Comparing with st (1) ....x>10^10 if 2.5^10 >2^14

or taking square root on both sides 2.5 ^5 >2^7 =>2.5^5 > 128

we know 25^2 is 625...hence 6.25x6.25x2.5 = 90 (approx)

Not greater

Hence A is sufficient
Case 2....X=2^35 is a fixed number

Hence ans is D
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by sureshbala » Fri Feb 06, 2009 2:14 pm
If you know the value of log 2 this is pretty simple.

The question asks is x > 10^10

i.e log x > 10

Statement 1:

x > 2^34

i.e log x > 34(0.3010)

Thus log x > 10.2

Hence from statement 1 we can conclude that x> 10^10

Statement 2:

Since the exact value of x is given we can conclude whether it will be more or less than that of 10^10.

Hence statement 2 alone is sufficient


It would definitely help if you can remember the log values of 2, 3, 5 and 7.

Remember these two simple rules of logarithms.

1. Let a and b both be greater than 1.

If log a > log b then a > b
If log a < log b then a < b

Converse is also true

2. Let 0 < a , b < 1.

If log a > log b then a < b
If log a < log b then a > b

Converse is also true

Even if you think along the lines of what DanaJ mentioned, you can avoid all the calculations to conclude 2^24 > 5^10.

log 2 = 0.3010 and log 5 = 0.6989

So log 2^24 = 24(0.3) = 7.2 and log 5^10 = 10(0.6989) = 6.98

Anyway, once you know how to apply logarithms there is no need to get to this stage and instead as I mentioned earlier you can easily conclude that statement 1 is sufficient
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by girish3131 » Mon Mar 15, 2010 4:37 am
Guys .... had the equation been x < 10^10


then of course ans wud have been D

but it is given x > 10^10

Acc to me ans is certainly E

long explanation... so can't writebut throwing a hint...

Take min x ie = 2^34

after solving

3225 * 3225 > 4096 * 4096

so here RHS is higher.... N we can still incease LHS because we have taken MINIMUM value of X

so can't say anything...

had sign reversed , we cud say something but not NOW...

same with EQUATION 2

Hence Ans is E


give one more try...

i hope u too find this one correct... no matter wat OA says....
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by Stuart@KaplanGMAT » Tue Mar 16, 2010 8:28 am
girish3131 wrote:Guys .... had the equation been x < 10^10


then of course ans wud have been D

but it is given x > 10^10

Acc to me ans is certainly E

long explanation... so can't writebut throwing a hint...

Take min x ie = 2^34

after solving

3225 * 3225 > 4096 * 4096

so here RHS is higher.... N we can still incease LHS because we have taken MINIMUM value of X

so can't say anything...

had sign reversed , we cud say something but not NOW...

same with EQUATION 2

Hence Ans is E


give one more try...

i hope u too find this one correct... no matter wat OA says....
Hi,

you seem to have reversed the question.

The question, not the equations, is:

Is x > 10^10?

So, we don't take it as a given that x is greater than 10^10 - that's what we're investigating.

It's also important to note that this is a yes/no question; we don't care if the answser is "yes" or "no", as long as we can get a definite answer to the question.

Even without doing math, we know that (2) is sufficient, since it provides an exact value for x. If we know the exact value for x, we can certainly determine whether it's bigger than 10^10.

Dana solved (1) and compared it to 10^10, proving that if (1) is true, x will always be greater than 10^10, so that statement is also sufficient alone.

Your mistake is a very common one in DS: instead of treating the question as a question, you treated it as a statement of reality. If we assume that the question is true statement, we've already answered it!
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by pims » Wed Jul 16, 2014 1:36 am
Rich?
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by GMATGuruNY » Wed Jul 16, 2014 2:55 am
Is x> 10^10?

1) x> 2^34
2) x= 2^35
To compare exponents, try to get SIMILAR BASES.
It is helpful to have memorized the powers of 2 up to 2¹�.
2¹� = 1024 ≈ 10³.

Statement 1: x > 2³�
2³� > 10¹�
2¹� * 2¹� * 2¹� * 2� > 10¹�
10³ * 10³ * 10³ * 16 > 10¹�
10� * 16 > 10� * 10
The lefthand side is greater than the righthand side.
Thus, x > 10¹�.
SUFFICIENT.

Statement 2: x = 2³�
Since the value of x is known, we can determine whether x > 10¹�.
SUFFICIENT.

The correct answer is D.
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by GMATinsight » Wed Jul 16, 2014 10:39 am
Is x> 10^10?

1) x> 2^34
2) x= 2^35
Another approach to the question could be

Step -1 Statement 2 gives a specific value of x which can certainly be compared with 10^10 and it is definitely sufficient to answer the question

Which leaves us with Options B and D only (A,C and E options can be ruled out)

Step - 2 x>2^34

Exponents can be compared when the powers on both sides are equal

2^34 = 2^4 x 2^10 x 2^10 x 2^10 = 16 x 1024 x 1024 x 1024 = 16 x 10^3 x 10^3 x 10^3 = 16 x 10^9
10^10 = 10 x 10^9

But 16 x 10^9 > 10 x 10^9
Therefore SUFFICIENT

Answer: Option D
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by Matt@VeritasPrep » Wed Jul 16, 2014 11:11 am
Mitch's approach is a great one; one other trick you can use here is the fact that 10·³ ≈ 2. (These values are very close; the one on the left is 10^(.3), if the superscript doesn't read properly on your browser.)

Since 10·³ ≈ 2, we know that 2³� ≈ (10·³)³� ≈ 10¹�·², which is greater than 10¹�. Hence x > 2³� > 10¹�.

Similarly, 2³� ≈ (10·³)³� ≈ 10¹¹, so 2³� > 10¹�.
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