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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