SOLUTION: Which of the following is the greatest?
a. {{{3^210}}}
b. {{{7^140}}}
c. {{{17^105}}}
d. {{{31^84}}}
Please give detailed solution.
(NOTE: Don't use logarithm. It has been
Algebra.Com
Question 977126: Which of the following is the greatest?
a.
b.
c.
d.
Please give detailed solution.
(NOTE: Don't use logarithm. It has been removed from syllabus, and I have to make my students understand without using log)
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Which of the following is the greatest?
a.
b.
c.
d.
---------------
3^210 = 9^105, obviously less than 17^105, so a is eliminated.
-------
Using a calculator:
7^140 = 2.059e118
17^105 = 1.57e129
31^84 = 1.88e125
-----
c wins again.
======================
You can change them all to the same base, eg, 7, but that would require either logs or a calculator.
----
eg,
7^140 = (2^2.8)^140 = 2^392
-------------
ooooooooooooooh, John's answer is cool.
I might have figured that out eventually.
I should have spotted that 3, 7, 17 & 31 are all 2^n plus or minus 1.
In short, his solution is elegant.
Probably with assistance from Satan, since we all know god is stupid.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
We can eliminate a for the reason Alan gave.
Then
^{140}\ =\ (2^3\ -\ 1)^{140}\ =\ 2^{420}\ -\ 140\cdot 2^{417}\ +\ ...)
By the binomial theorem, and
^{105}\ =\ (2^4\ -\ 1)^{105}\ =\ 2^{420}\ +\ 105\cdot 2^{416}\ +\ ...)
Notice that in the expansion of ^{140})
, the signs alternate, but in the expansion of ^{105})
the signs are all positive. Therefore
Similarly, the first term of ^{84})
will be 
and the terms will alternate, hence
Therefore, answer c, as Alan's calculator confirms.
John
My calculator said it, I believe it, that settles it
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