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As in so many of these problems, the first task is careful reading. Note that x is not itself the integer (except in the case of 1/1); x is the reciprocal of an integer. You might even make a quick table corresponding to the values in the answer choices: Now, just keep adding columns. Add one for y, defined as –x^2: Notice that according to PEMDAS, you apply the squaring operation (Exponent) before the negative sign (Subtraction).
Add one more column for x^y. For clarity, we’ll use the caret symbol (^) to indicate exponents. Notice that the negative sign undoes the reciprocal in x, leaving you with an integer base. Finally, you are left with the task: which of those numbers in the last column is largest? Remember, fractional exponents are roots, so the second number is the fourth root of 2, and so on.
You can quickly eliminate 1, since the positive roots (to any degree) of any integer greater than 1 are always greater than 1. For instance, what is the 25th root of 5? It must be a number bigger than 1, even if only slightly, because that number times itself 25 times in all produces 5. So you can eliminate A.
Let’s compare the second and fourth numbers, because the base of 4 can be rewritten as 2^2. The fourth answer becomes then \((2^2)^{(\frac{1}{16})}=2^{(\frac{1}{8})}\), which is less than 2^(1/4). The number that solves z^8 = 2 is smaller than the one that solves z^4=2. So the answer can’t be D.
What about B versus C? Raise both numbers to the 36th power, so that we eliminate fractional exponents.
2^(1/4)^36=2^9
3^(1/9)^36=3^4
Which of the results is bigger? 2^9=512, while 3^4= only 81. So the original numbers must be in that same order of size; B is bigger than C. C is out.
A similar argument takes out E (raise both B and E to the 100th power):
If x is the reciprocal of a positive integer, then the maximum value o
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Updated on: 01 May 2015, 05:27
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Answer is B. [edited from below]
As corrected below: I originally bit on answer choice E the 25th root of 5 as soon as I knew it was greater than 1. However the 4th root of 2 is greater than the 25th root of 5.
Take the reciprocal of 5 = 1/5 = x
y = -(x^2) therefore y = -(1/25)
x^y = (1/5)^ -(1/25) = 25th square root of 5. To solve negative exponents, take the reciprocal of the base, and apply the exponent as its absolute value. In this case it results: (1/5)^ -(1/25) = 5^(1/25). The 25th square root of 5 is small, but it's greater than 1.
Bunuel, is there any good rule of thumb for comparing the value of square roots?
Originally posted by ak1802 on 29 Apr 2015, 06:50.
Last edited by ak1802 on 01 May 2015, 05:27, edited 2 times in total.
If x is the reciprocal of a positive integer, then the maximum value o
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29 Apr 2015, 06:56
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ak1802 wrote:
Answer is E.
Take the reciprocal of 5 = 1/5 = x
y = -(x^2) therefore y = -(1/25)
x^y = (1/5)^ -(1/25) = 25th square root of 5. To solve negative exponents, take the reciprocal of the base, and apply the exponent as its absolute value. In this case it results: (1/5)^ -(1/25) = 5^(1/25). The 25th square root of 5 is small, but it's greater than 1 and it's also greater than all other answer choices.
Thanks for reprimand. But in this case I think the correct answer is B because 4th root from 2 will be bigger than 25th root from 5.
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29 Apr 2015, 07:16
Harley1980 wrote:
ak1802 wrote:
Answer is E.
Take the reciprocal of 5 = 1/5 = x
y = -(x^2) therefore y = -(1/25)
x^y = (1/5)^ -(1/25) = 25th square root of 5. To solve negative exponents, take the reciprocal of the base, and apply the exponent as its absolute value. In this case it results: (1/5)^ -(1/25) = 5^(1/25). The 25th square root of 5 is small, but it's greater than 1 and it's also greater than all other answer choices.
Thanks for reprimand. But in this case I think the correct answer is B because 4th root from 2 will be bigger than 25th root from 5.
You're correct. I just checked the math in Excel.
Oversight on my part, I jumped to the conclusion of 5 once I realized it is greater than 1 without out considering the 4th root of 2.
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30 Apr 2015, 11:13
When we input answer choices into the function we end up with the comparison between these values A)\(1\) B)\(2^\frac{1}{4}\) C)\(3^\frac{1}{9}\) D)\(4^\frac{1}{16} = 2^\frac{1}{8}\) E)\(5^\frac{1}{25}\) Obviously A is out, B > D (D is out) B > C because \(2^\frac{9}{4} = 4*2^\frac{1}{4} > 3\) coz \(2^\frac{1}{4} > 1\) thus C is out E = \(5^\frac{1}{25} < 8^\frac{1}{25} = 2^\frac{3}{25} < 2^\frac{3}{24} = 2^\frac{1}{8} < 2^\frac{1}{4}\) so E < D < B, thus E is out B left, B is the answer
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05 Nov 2016, 07:00
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Bunuel wrote:
If x is the reciprocal of a positive integer, then the maximum value of x^y, where y = –x^2, is achieved when x is the reciprocal of
A. 1 B. 2 C. 3 D. 4 E. 5
\(x=\frac{1}{n}\), and we have \((\frac{1}{n})^y\). \(y=-x^2\), => \(x^y\) =\(\sqrt[n^2]{n}\)
We can plug in different values and compare them. It will definitely take some time. But do we need to do this. The answer is no.
If we look carefully at this expression we’ll notice that \(n^2\) is growing much faster than \(n\). Saying this in other words, the bigger the value of \(n\), the smaller the value of the expression \(x^y\).
In fact we can take \(2\) from the start. \(n=1\) is the minimum value.
By noticing that \(\sqrt[n^2]{n}\) will always be bigger than 1 for n>1, we got only suitable choice for max value of \(x^y\) as \(\sqrt[4]{2}\). For all other \(n>2\), the value of \(\sqrt[n^2]{n}\) will only decrease and will be closing to \(1\). This will save a lot of time.
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15 Mar 2018, 09:15
I think taking logs helps in these kinds of problems.
The effective equation is --> (1/a^2)*log(a), where a belongs to +I. This is maximum for 2. You since the expression value for a =2 is greater than a= 4.
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10 Sep 2019, 10:56
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