Comparison Mania

Algebra Level 3

Let

$a = 2^{147}$

$b = 3^{105}$

$c = 4^{63}$

$d = 5^{63}$

$e = 7^{42}$

Arrange the values in descending order. Give your answer in the following form:

If you get $C > A > D > E > B$ , your answer would be 31452.

If you get $A > E > B > D > C$ , your answer would be 15243.

(I did not come up with this question. This is from the entrance exam of the Computer POSN camp in Thailand.)

The answer is 21435.

3 solutions

Pi Han Goh
Apr 1, 2014

Hint: for positive real numbers, $V,W,X,Y,Z$ are all greater than $1$ , if $V^{21}<W^{21}<X^{21}<Y^{21}<Z^{21}$ , then $V<W<X<Y<Z$

Solved the same way......

Satvik Golechha - 7 years, 2 months ago

I solved using logarithms. Is that not correct?

Krishna Ar - 7 years, 2 months ago

It's still correct. But it's cooler if you didn't....

Pi Han Goh - 7 years, 2 months ago

Maharnab Mitra
May 4, 2014

Take log for each one

Then, $105log3>147log2>63log5>63log4>42log7$ which is also the order of the numbers.

You can get the value of log from a table or calculator.

Salah Amer
Apr 24, 2014

b=3^106=9^53>8^53 Then b>2^3^53 Then b>2^159 Then b > 2^147 Then b>a Also a=2^147=(2^63) (2^63) (2^21)=(4^63) (2^(63/3))=(4^63) (2^(1/3))^63 And d=5^63=(4^63) ((1.25)^63) But (2^(1/3))^63>(1.25)^63 Then a>d Also d>c (it is clear) Also e=7^42=(4^42) ((7/4)^42)=(4^42) ((7/4)^21) ((7/4)^21)=(4^42) ((4 7/16)^21) ((7/4)^21) =(4^42) (4^21) ((7/16)^21) ((7/4)^21)=(4^63)*((49/64)^21) Then e<4^63 Then c>e

Comparison Mania

The answer is 21435.

3 solutions

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