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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