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Re: Which of the following is the largest? [#permalink]

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25 Nov 2013, 03:32

gmatprav wrote:

(E)\(11^{5.1}\) approximately \(10^5\)

You can't approximate\(11^{5.1}\) to \(10^5\). This is not approximation, it is just lowering the actual value.If anything,it might be approximated to \(11^5\)

gmatprav wrote:

B) approximated to \(10^9\) C) \(\frac{10^{11}}{2^{11}}\) = \(\frac{10^{11}}{1024}\) = \(\frac{10^{11}}{10^3}\) = \(10^8\) E) \(10^5\)

\(2^{11}\) not 1024, it is 2048.Nonetheless, it doesn't hurt as because if anything it lowers the value even further.

IMO,your methodology for eliminating A and D is perfect, otherwise you stretched the scope for approximating. Worked this time, might not work in some close case encounters.
_________________

Re: Which of the following is the largest? [#permalink]

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25 Nov 2013, 04:19

1

This post received KUDOS

mau5 wrote:

gmatprav wrote:

(E)\(11^{5.1}\) approximately \(10^5\)

You can't approximate\(11^{5.1}\) to \(10^5\). This is not approximation, it is just lowering the actual value.If anything,it might be approximated to \(11^5\)

gmatprav wrote:

B) approximated to \(10^9\) C) \(\frac{10^{11}}{2^{11}}\) = \(\frac{10^{11}}{1024}\) = \(\frac{10^{11}}{10^3}\) = \(10^8\) E) \(10^5\)

\(2^{11}\) not 1024, it is 2048.Nonetheless, it doesn't hurt as because if anything it lowers the value even further.

IMO,your methodology for eliminating A and D is perfect, otherwise you stretched the scope for approximating. Worked this time, might not work in some close case encounters.

Thanks for pointing out the approximation mistake. One new thing I realized with this post, hope this is correct: For positive integers m & n, \(m^n\) > \((m+1)^{n-1}\) Using this and avoiding excess approximation.

B is \(9^9\) and C is \(\frac{10^8}{2}\) and E is \(11^5\).

\(9^9 > 10^8 > 10^8/2\) and \(11^5\) is significantly lower than \(10^8/2\). we arrive at B as solution.
_________________

We know that \(2^{3}<3^{2} \to\) Raising both sides to the power of 9.1, we have \(3^{18.2}>2^{27.3}\)

Again, we know that\(3^{2}>7 \to\) Raising both sides to the power of 9.1, we have \(3^{18.2}>7^{9.1}\)

Also,\(5^3>11^2\) ; Raising both sides to the power of 3.7, we have \(5^{3*3.7}>11^{2*3.7} \to 5^{11.1}>11^{7.4}\). Thus, \(5^{11.1}\) has to be greater than \(11^{5.1}\).

Finally, \(3^3>5^2 \to\) raising on both sides to the power of 6, we have \(3^{18}>5^{12}\). Thus, \(3^{18.1}\) will always be greater than \(5^{11.1}\).

First look for some commonality - either power or base. I see that 27.3 is 3*9.1 and 18.2 is 2*9.1. So (A) \(2^{27.3} = 8^{9.1}\)

(B) \(3^{18.2} = 9^{9.1}\)

(D)\(7^{9.1}\)

Out of (A), (B) and (D), (B) is the largest.

We can easily compare \((C) 5^{11}\) with \((E) 11^{5}\). Out of a^b and b^a, greater is usually the one with the smaller base but higher power. Only in early powers of 2 is this pattern not followed. \(2^3 < 3^2, 2^4 = 4^2\) but \(2^5 > 5^2\) and the gap keeps widening. Similarly, \(3^4 > 4^3\) and the gap keeps widening. So \(5^{11}\) will be greater than \(11^{5}\). So (C) is greater. Now we just need to compare (B) with (C)

(B) \(3^{18} = 27^6\) (C) \(5^{11} = 25^{5.5}\)

(B) has a greater base and a greater power so it is obviously greater.

Answer (B)

Note:

In case you want to compare (C) and (E) in another way, (C) \(5^{11.1} = 125^{3.7}\) (E)\(11^{5.1} = 121^{2.5}\)

(C) has greater base and power so is greater.
_________________

Re: Which of the following is the largest? [#permalink]

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24 Aug 2016, 21:17

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Re: Which of the following is the largest? [#permalink]

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21 Sep 2017, 00:25

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This post was BOOKMARKED

OFFICIAL SOLUTION FROM MANHATTAN

When comparing numbers raised to different powers, the first method is typically to try to break the bases down and find similarities. Here, though, the bases are all prime numbers already, and they’re all different! Now what?

As always, start with the most annoying feature: those decimals in the exponents. If you raise every number to the 10th power, you’ll still preserve the relative order of the 5 numbers. All you care about is finding the largest, and because you’re starting with 5 positive bases, the largest number will remain the largest even after raising each to the 10th power.

(A) 2^273 (B) 3^182 (C) 5^111 (D) 7^91 (E) 11^51

So, let’s see, what’s 2 raised to the 273 power? 2, 4, 8, 16, 32, 64, … just kidding.

Check out the exponents: are there any similarities? Any common factors?

Yes! 273 and 182 are both multiples of 91. Take the 91st root of (A), (B), and (D) so that you can compare just those three. Taking the 91st root is the same as dividing all three by 91:

Excellent! Among these three, then, answer (B) is the largest; you can eliminate (A) and (D).

As for the other two choices, compare them one at a time to one of the three you've already compared, looking for benchmarks. Take (C): 5^111. Are there any small powers of 5 close to a power of 2, 3, or 7? Yes: 52 = 25, while 33 = 27.

How do we use that? Well, 33 > 52. Scale up by taking both sides to the 60th power: 3^180 > 5^120. This inequality slips nicely between the relevant answer choices. If 3^180 > 5^120 then the larger 3^182 must be bigger than the smaller 5^111. Eliminate answer (C).

Try this tactic again with answer (E): 11^2 = 121 and 2^7 = 128. In this case, 2^7 > 112. The power of 2 in answer (A) is 273, which divided by 7 yields 39. So take both sides of the inequality to the 39th power: 2^273 > 1178, which is larger than 11^51. Answer (E) is not the largest either!

The correct answer is (B).
_________________

we shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender!

Which of the following is the largest? [#permalink]

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22 Sep 2017, 11:18

I use a shortcut for such kind of questions, we need to use natural logarithm , not for calculations but for its properties Exponential function is an inverse function of logarithm function And there a common property for all type logarithm but for the simplicity sake lets stick to natural only ln(x^10)= 10*ln(x) Then it's easy to memorize 6 numbers: ln 2 ~ 0.7 (then for ln 4 just multiply 0.7 by 2, for ln 8 by 3 and so on, exponential growth) ln 3 ~ 1.1 ln 5 ~ 1.6 ln 6 ~ 1.8 ln 7 ~ 1.95 ln 10 ~ 2.3

So for example: (A) 2^27.3 => 27.3*ln 2 => 0.7*27.3 (B) 3^18.2 => 18.2*ln 3 => 18.2*1.1 (C) 5^11.1 => 11.1*ln 5 => 11.1*1.6 (D) 7^9.1 => 9.1*ln 7 => 9.1*1.95 (E) 11^5.1 => 5.1*ln 11 you can replace it for 6*ln 10 don't bother yourself a lot=> 6*2.3

I think its obvious, we have our champ B 'cause A gonna be < 20 as well as the rest of them, and B above If you memorize those 6 numbers and understand the approach you can solve that kind of question in 30-40 seconds

First look for some commonality - either power or base. I see that 27.3 is 3*9.1 and 18.2 is 2*9.1. So (A) \(2^{27.3} = 8^{9.1}\)

(B) \(3^{18.2} = 9^{9.1}\)

(D)\(7^{9.1}\)

Out of (A), (B) and (D), (B) is the largest.

We can easily compare \((C) 5^{11}\) with \((E) 11^{5}\). Out of a^b and b^a, greater is usually the one with the smaller base but higher power. Only in early powers of 2 is this pattern not followed. \(2^3 < 3^2, 2^4 = 4^2\) but \(2^5 > 5^2\) and the gap keeps widening. Similarly, \(3^4 > 4^3\) and the gap keeps widening. So \(5^{11}\) will be greater than \(11^{5}\). So (C) is greater. Now we just need to compare (B) with (C)

(B) \(3^{18} = 27^6\) (C) \(5^{11} = 25^{5.5}\)

(B) has a greater base and a greater power so it is obviously greater.

Answer (B)

Note:

In case you want to compare (C) and (E) in another way, (C) \(5^{11.1} = 125^{3.7}\) (E)\(11^{5.1} = 121^{2.5}\)

(C) has greater base and power so is greater.

Responding to a pm:

Quote:

Hi Karishma Please how did you calculate these? (B) \(3^{18} = 27^6\) (C) \(5^{11} = 25^{5.5}\)