Exponential

[Anindya Madhudor]

Is x > 10^10?
i. x> 2^34
ii. x=2^35

[Brent@GMATPrepNow]

Is x > 10^10?
i. x> 2^34
ii. x=2^35

Target question: Is x > 10^10?

Statement 2: x = 2^35
Notice that I'm looking at statement 2 first. Why?
Well, this statement tells me the exact value of x.
So, if I wanted to, I could evaluate 2^35, and then determine whether or not x is greater 10^10
So, since I could use statement 2 to answer the target question with certainty, statement 2 is SUFFICIENT

Important: Now that I know statement 2 is sufficient, the correct answer must be either B or D. Great! After 5 seconds work, I have a 50% chance of guessing correctly.

Statement 1: x > 2^34
This one is much trickier. Here's how I'd approach the question.

Edit: This is a lengthy solution. I realized a faster approach, which can be found in the post after this one.

I basically need to compare 2^34 with 10^10.
If 2^34 is less than 10^10, then x could be greater than or less than 10^10
If 2^34 is greater than 10^10, then x must be greater than 10^10

Important: Notice that 2^34 = (2^3.4)^10
So rather than compare (2^3.4)^10 with 10^10, we can compare 2^3.4 with 10
Which is greater?

A calculator would be nice, but no such luck.
Now I do happen to know that sqrt(2) = 1.4 (approximately), and this will help us get a feel for the value of 2^3.4

Aside: I recommend that students memorize the following roots:
sqrt(2) = 1.4 (approximately)
sqrt(3) = 1.7 (approximately)
sqrt(5) = 2.2 (approximately)
These can come in handy at times . . . like now!

So, let's examine 2^3.5 (which is kind of close to 2^3.4)

2^3.5 = (2^3)(2^0.5)
= (8)(sqrt2) [since k^0.5 = sqrtk]
= (8)(1.4) ...approximately
= 11.2 approximately

So, I know that 2^3.5 = 11.2 (approx)
From this, what can we conclude about 2^3.4?
Well, we might use a bit of number sense to conclude (correctly) that 2^3.4 is greater than 10.
So, if 2^3.4 > 10, we can be certain that (2^3.4)^10 > 10^10, which means 2^34 > 10^10, which means x must be greater than 10^10

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Answer = D

Cheers,
Brent

[Brent@GMATPrepNow]

Is x > 10^10?
i. x> 2^34
ii. x=2^35

After thinking about it, here's a better approach to statement 1:

Target question: Is x > 10^10?

Statement 1: x > 2^34
As I mentioned in my first solution, we basically need to compare 2^34 with 10^10.
Now notice that 2^34 = (2^10)(2^24)
Also notice that 10^10 = (2^10)(5^10)

So, if we divide both quantities by 2^10, we can see that we need to compare 2^24 with 5^10

Now notice that:
2^24 = (2^12)^2
and 5^10 = (5^5)^2

So, if we find the square root of both quantities , we can see that we need to compare 2^12 with 5^5
This is pretty manageable.

2^12 = (2^6)(2^6)
= (64)(64)
= 3600+

5^5 = (5^4)(5)
= (625)(5)
= 3100 (approx)

So, since 2^12 > 5^5, we can be certain that 2^24 > 5^10, which means 2^34 > 10^10, which means x must be greater than 10^10

Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Answer = D

Cheers,
Brent

[Anindya Madhudor]

Thank you, Brent. Really nice way to solve the problem. Appreciate your help.

[puneetkhurana2000]

Hi,

Before we start just to let you guys know that this may not be the best method to solve exponents' inequalities if you are not familiar with logs.

Statement 2) x=2^35

Sufficient as we can easily deduce the value of x and find whether it is larger or smaller than 10^10.

Statement 1) x > 2^34

Taking log on both sides with base 2 will result into ... log(base 2)x > log(base 2)2^34
So, LHS > 34 as log(base 2)2 = 1 ...using formula (log(base a)a = 1)

Now let us suppose y = 10^10
Taking log on both sides with base 2 will result into ... log(base 2)y = log(base 2)10^10
So, LHS = 10*log(base 2)10 and log(base 2)10 is between 3 and 4 and closer to 3 as 2^3 = 8 and 2^4 = 16, so LHS = 10*(a quantity closer to 3) = 30(approx)

So, 2^34 is > than 10^10 because a number greater than 34 (assumed x above) must be greater than a
number closer to 30 (assumed y above).

Sufficient!!!

Answer D

[GMATGuruNY]

Is x> 10^10?

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

Quick solution to verify A?

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.