GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If d = 2^64, and d^d = 2^p, p = ?
[#permalink]
23 Nov 2021, 04:47
23 Nov 2021, 04:47
Expert Reply
00:00
Question Stats:
66% (01:16) correct
33% (02:08) wrong based on 3 sessions
Hide Show timer Statistics
If \(d=2^{64}\), and \(d^d=2^p\), \(p=?\)
(A) \(2^{68}\)
(B) \(2^{70}\)
(C) \(2^{96}\)
(D) \(2^{128}\)
(E) \(2^{256}\)
(A) \(2^{68}\)
(B) \(2^{70}\)
(C) \(2^{96}\)
(D) \(2^{128}\)
(E) \(2^{256}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If d = 2^64, and d^d = 2^p, p = ?
[#permalink]
23 Nov 2021, 05:24
23 Nov 2021, 05:24
Carcass wrote:
If \(d=2^{64}\), and \(d^d=2^p\), \(p=?\)
(A) \(2^{68}\)
(B) \(2^{70}\)
(C) \(2^{96}\)
(D) \(2^{128}\)
(E) \(2^{256}\)
(A) \(2^{68}\)
(B) \(2^{70}\)
(C) \(2^{96}\)
(D) \(2^{128}\)
(E) \(2^{256}\)
Let's first determine an equivalent value of \(d^d\) by replacing \(d\) with \(2^{64}\)
We get: \(d^d = (2^{64})^{(2^{64})}\)
Recall the Power of a Power rule: \((b^x)^y = b^{xy}\)
Let's compare \((2^{64})^{(2^{64})}\) with \((b^x)^y\)
We can see that \(b=2\), \(x = 64\) and \(y = 2^{64}\)
This means we can calculate the product \(xy\) as follows: \(xy = (64)(2^{64})\)
Rewrite \(64\) as \(2^6\) to get: \(xy = (2^6)(2^{64})\)
Apply the Product law to get: \(xy = 2^{70}\)
Now that we know \(b=2\) and \(xy = 2^{70}\), we can plug these values into the Power of a Power rule to get: \(b^{xy}=2^{(2^{70})}\)
So, \(d^d=2^p\) becomes \(2^{(2^{70})}=2^p\), which means \(p=2^{70}\)
Answer: B
RELATED VIDEO
gmatclubot
Re: If d = 2^64, and d^d = 2^p, p = ? [#permalink]
23 Nov 2021, 05:24
Moderators:
