Which of the following has the greatest value? Try to answer without using a calculator!
Note:
x
y
denotes the
x
th
power root of
y
:
i.e.
x
y
=
y
1
/
x
.
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I "cheated" and looked at the function y = x x 1 , finding its derivative to be
y ′ = x x 1 − 2 ( 1 − ln ( x ) ) .
The critical point ln ( x ) = 1 ⟹ x = e yields a maximum for y of e e 1 , and the nearest integer to e is 3 , making it a pretty good bet.
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That was my original solution, before I found the more elementary one that I posted. If you look at the logarithm, y = x ln x , the derivative will be easier to work with.
To improve the "pretty good bet", note that for 2 < x < 4 , x x > 2 2 = 4 4 because the function is concave on that entire interval.
2 root 2 and 4 root 4 are the same ;)
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That is why I wrote:
(Or save yourself some work by noticing 2 2 = 4 4 .)
Let f ( x ) = x x 1 We need to compare f ( 2 ) , f ( 3 ) , f ( 4 ) , f ( 5 )
ln ( f ( x ) ) = x ln ( x ) Differentiating, f ( x ) f ′ ( x ) = x 2 x / x − ln ( x )
For maximum, f ′ ( x ) = 0
1 − ln ( x ) = 0
So maximum value of f ( x ) is at x = e
So, we can for sure say that f(3)>f(4)>f(5) so we need to only compare f(2) and f(3).
If you do not want to compare, just see that 3 is much closer to e than 2. This does not work always so better to check but the answer is 3
You can compare the f(2) and f(3) cases without using a calculator as follows -
The cube of the square root of two is 2 2 which is less than 2 × 1 . 5 = 3 . So the cube root of 3 must be greater than the square root of two.
f'(x) =0 is a critical point, would want to include that f''(e)=-1/e^3 to complete the justification of a maximum. Or a similar 1st derivative test argument would do.
Dude, f(2) = f(4). The heck are you doing?
First, convert each radical to its fractional exponent form:
2
2
=
2
2
1
3
3
=
3
3
1
4
4
=
4
4
1
5
5
=
5
5
1
Now compare 2 2 1 and 4 4 1 . With some simplification you can notice that they are equal.
2 2 1 = 4 4 1
Now compare 3 3 1 and 5 5 1 . Notice that 3 3 and 5 2 are almost equal. We can use this fact to compare the two values.
Let x = 3 3 1 and y = 5 5 1 . This means x 9 = 3 3 and y 1 0 = 5 2 . As x 9 and y 1 0 are now almost equal, it is clear that x > y as x has a smaller exponent. Therefore,
3 3 1 > 5 5 1 .
Now you could compare 2 2 1 and 3 3 1 , however each answer choice only contains one option so it is clear that the answer must be 3 3 1 as 2 2 1 = 4 4 1 .
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First, compare 2 2 and 3 3 by raising each to the sixth power: ( 2 2 ) 6 = 2 3 = 8 ; ( 3 3 ) 6 = 3 2 = 9 ; since 8 < 9 , 2 2 < 3 3 . Next, compare 3 3 and 4 4 by raising each to the twelfth power: ( 3 3 ) 1 2 = 3 4 = 8 1 ; ( 4 4 ) 1 2 = 4 3 = 6 4 ; since 8 1 > 6 4 , 3 3 > 4 4 . (Or save yourself some work by noticing 2 2 = 4 4 .)
Finally, compare 3 3 and 5 5 by raising each to the fifteenth power: ( 3 3 ) 1 5 = 3 5 = 2 4 3 ; ( 5 5 ) 1 5 = 5 3 = 1 2 5 ; since 2 4 3 > 1 2 5 , 3 3 > 5 5 .
Thus 3 3 is the biggest of the bunch.