Which of the following is the largest?

Question

(A)  2^{27.3}

(B)  3^{18.2}

(C)  5^{11.1}

(D) 7^{9.1}

(E) 11^{5.1}

Fresh from Manhattan GMAT (Challenge of the week)

KarishmaB · Accepted Answer

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.

mau5 · Answer

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

Again, we know that 3^{2}>7 	o  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} 	o 5^{11.1}>11^{7.4} . Thus,  5^{11.1}  has to be greater than  11^{5.1} .

Finally,  3^3>5^2 	o  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} .

B.

gmatprav · Answer

Each of these can be represented as below: (A) 2^{27.3} = (2)^{3*9.1} = (2^3)^{9.1} = 8^{9.1} (B) 3^{18.2} = (3)^{2*9.1} = (3^2)^{9.1} = 9^{9.1} (C) 5^{11.1} = (\frac{10}{2})^{11.1} = \frac{10^{11.1}}{2^{11.1}} (D) 7^{9.1} = 7^{9.1} (E) 11^{5.1} approximately 10^5 Among A,B & D we have B as clear winner. Now choices are down to B,C & E From here I did some approximation 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 Among these B is the largest. PS: Typing this solution took longer than solving it

mau5 · Answer

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

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.

gmatprav · Answer

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.

bkpolymers1617 · Answer

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:
 
(A) 2^273 = 23 = 8
(B) 3^182 = 32 = 9
(D) 7^91 = 71 = 7
 
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).

cbh · Answer

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

KarishmaB · Answer

Responding to a pm: 3^{18} = 3^{3*6} = (3^3)^6 = 27^6 5^{11} = 5^{2*5.5} = (5^2)^{5.5} = 25^{5.5} Here is a video on exponents that you will find useful: https://youtu.be/ibDqnatAMG8

PolarisTestPrep · Answer

Hey!

The GMAT really loves tricks, and this is a perfect example.

If you look at this and start calculating -- stop. The GMAT will never ask you to do so many complex calculations. The GMAT is a test of management above all else and managers are not mathematicians -- they are efficient problem solvers. So the GMAT will give you a problem like this that will make your first instinct go, "If I do these calculations fast enough maybe I will get the answer", or maybe your first instinct says, "God, why is this happening to me?", and that is how they like to trip you up.

The instant you get an urge to do a complex calculation or lose all hope, this is the trigger you should learn to stop immediately and ask yourself, "What is the trick here? What is the sneaky shortcut around the complex, seemingly impossible calculations here?" What they are really testing is whether you see the clever and fast way through the problem or not, not whether you can do the complex math.

When you see exponents with decimals your brain should rightfully explode. The difference that will lead to success on the test is that this is not a call to do the hard math but a call to try and spot the trick. Approach all practice problems with this in mind and you can develop an eye for the tricks. Learning the math will get you a decent baseline score, learning the tricks will get you 700+.

When you see something as seemingly intractable as this, you should start to look for alternative ways to solve, and hopefully your eye catches that 9.1, 18.2, and 27.3 are multiples of the same number. You simply must develop an eye for such things -- look for similarities between elements, ways to break down complicated things into simpler component parts, etc. If you approach the difficult problems with this mindset rather than trying to math your way through it, you will eventually develop this eye and this instinct to ask, "What's the catch here?" Because that is what this test, and particularly at the higher level, is really all about.

So if you see these multiples, you can break them down into smaller parts:

A.  2^{27.3}  is  2^{(3)(9.1)}  which is  8^{9.1}  because  2^{3} = 8 
B.  3^{18.2}  is  3^{(2)(9.1)}  which is  9^{9.1}  because  3^{2} = 9 
D.  7^{9.1}

We are comparing A.  8^{9.1} , B.  9^{9.1} , and D.  7^{9.1}

They all have the same exponent now so we can truly compare them

B.  9^{9.1}  is the largest of the three as it is the largest base number raised to the same exponent

What about C and E?

First, determine which of the two is larger to compare with B.

You could notice that half of 11.1 is close to 5.1

Compare C.  5^{(2)(5.55)}  which is  25^{5.55}  with E.  11^{5.1}

C.  25^{5.55}  vs. E.  11^{5.1}

The key is to get the two elements as similar to each other as possible to make it easier to compare them. Here, we got the exponents near enough to each other that the answer becomes obvious.

C has both the larger base and the larger exponent, so it must be the larger number.

That is one way to do it, but there are other tricks and shortcuts to use here. As posted above, you could know that, generally,  a^{b}  will be larger than  b^{a}  when  b>{a} . Learning these kinds of math rules will really go a long way in speeding up your ability to answer questions and catch shortcuts. The good thing is there usually are several shortcuts available depending on what you know and what you spot so if you learn as much as possible and practice being sneaky, you're that much more likely to spot a clever way through a problem.

In the end we have B.  3^{18.2}  vs. C.  5^{11.1}

If the answer is not obvious, again look for ways to manipulate and rewrite to make the two elements more similar and easier to compare.

3^{18.2}  can be rewritten as  3^{(3)(6.6-)}  which is   27^{6.6-}  and  5^{11.1}  can be rewritten as  5^{(2)(5.55)}  as before which is  25^{5.55}

This gets the bases closer to each other, making it that much easier to compare the two numbers. The reason I don't calculate the actual decimal for  27^{6.6-}  is it doesn't matter -- look at the rough estimation and see if it gives you enough information to answer the question, every second is precious on the GMAT.

B.  27^{6.6-}  vs. C.  25^{5.55}

Both the base and the exponent are larger in B, regardless of what the hundredths number of the exponent is

(B) is the largest number of the five and the answer to the question

Crytiocanalyst · Answer

The key to solving the problem is getting the approximation correct

(A) 2^27.3
=>2^26 = (2^13)^2
=>Approx(8000)^2

(B) 3^18.2
=>(3^9)^2=(27^3)^2
=>Approx(27000)^2

(C) 5^11.1
=>(5^5)^2 =(125*25)^2
=>approx(3000)^2

(D) 7^9.1
=>(343)^2

(E) 11^5.1
=>(121)^2

Clearly B is the greastest therefore 
IMO B

Which of the following is the largest?

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Which of the following is the largest?

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