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Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500

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Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500  [#permalink]

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New post 26 Nov 2018, 23:44
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

70% (01:59) correct 30% (02:16) wrong based on 44 sessions

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Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500  [#permalink]

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New post 27 Nov 2018, 00:04
Bunuel wrote:
Is \(2^x > 75\) ?


(1) \(3^x > 50\)

(2) \(2^x*3^x > 1500\)


We should be familiar with the powers of 2 to know that 2^6 = 64 and 2^7 = 128.
So, (assuming x is an integer) we need to know if x > 6. Even if it isn't, this is a reasonable first approximation.

We'll look for statements that give us this information, a Logical approach.

(1) 3^3 = 27 and 3^4 = 81, so this tells us that x > 3. But is it larger than 6?
Insufficient.

(2) Again, we'll look for the edge of the range. We can either guess numbers or first use prime factorization: 1500 = 15 * 100 = 5*3*10^2 = 5*3*(5*2)^2 = (2^2)(3)(5^3). S

so if x = 6 then we can compare (2^6)(3^6) with (2^2)(3)(5^3). Canceling out (2^2)(3) from both gives (2^4)(3^5) = 16*243 on the left and 5^3 = 125 on the right.
So x = 6 is definitely enough. Since it can also be larger, this is once again insufficient.

Combining adds no new information (as all we have are inequalities and since x = 6 and x > 6 are good for both inequalities, they will be for their combination)

(E) is our answer.
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Re: Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500  [#permalink]

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New post 27 Nov 2018, 02:26
Can you please explaim stmt b
I am not clear post the prime factorization

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Re: Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500  [#permalink]

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New post 27 Nov 2018, 03:57
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saurabh9gupta wrote:
Can you please explaim stmt b
I am not clear post the prime factorization

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Hi,

Basically, if you plug x = 6 into statement (2) you get 2^x * 3^x = 2^6 * 3^6 which is larger than 1500. So, based on (2), x could equal 6 in which case 2^x < 75 and the answer is NO. But since x could also equal 7, and 2^7 is larger than 75 then this gives a YES. Two different answers - insufficient!

In the solution above, I used prime factorization because it gave me easier numbers to compare. Instead of actually calculating 2^6, 3^6 and multiplying, I could cancel out 4*3 = 12 from both 2^6*3^6 and 1500 which then let me make an easier calculation of 2^4 * 3^5. Doesn't really matter too much, just a question of how you can do the math as fast as possible. Directly approximating also works: 2^6 *3^6 = 64 * 27 * 27 = about 60 * 30 * 30 = much more than 1500, which is also perfectly fine.
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Re: Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500  [#permalink]

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New post 27 Nov 2018, 04:09
2^x > 75?

We need to find the value of x to determine whether 2^x > 75. So what is the value of x?

St 1: 3^x > 50

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243


In order to satisfy the statement, x should be greater than or equal to 4 (x>=4)
Now, lets apply this to the main question. 2^4 = 16, 2^5 = 32, 2^6 = 64, 2^7 = 128.
Here, 2^7 > 75 but 2^4 < 75. Hence the statement is insufficient .

St 2: 2^x ∗ 3^x > 1500

In order to satisfy this statement, x should be 5 or greater than 5. (2^5 ∗ 3^5 = 7776)
Back to the main question, 2^5 = 32. Is 32>75(No) but 2^7=128 which is greater than 75.
Therefore, the statement is clearly insufficient.

St 1+ St 2:

We have no additional information.
So far, we know from statement 1 and statement 2 that x>= 4 but that will not help us to get a unique answer. Therefore C is clearly insufficient too.

We are left with (E) & that's the answer IMO.

Thank you
Arjun
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Re: Is 2^x > 75 ? (1) 3^x > 50 (2) 2^x*3^x > 1500 &nbs [#permalink] 27 Nov 2018, 04:09
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