math_kp partially solved a similar problem last night:
Take the derivative of the function y = x ^ (1/x),
y' = y / x^2 * (1 - ln(x))
and notice it's 0 at x = e, and negative for x > e.
Since a = 89 and b=90, both a,b > e, and a
a^(1/a) > b^(1/b) or
a^b > b^a
89^90 > 90^89
2006-10-11 01:48:19
·
answer #1
·
answered by Joe C 3
·
0⤊
0⤋
A-B =C
if C>0 than A > B
if C<0 than A
using calculator
89^90 - 90^89 =X , X>0
90^89-89^90 =X , X<0
so conclusion
89^90 > 90^89
2006-10-11 08:57:41
·
answer #2
·
answered by safrodin 3
·
0⤊
0⤋
You dont need mathematical proof - they are just numbers.
89^90 = 2.78676661 Ã 10^175
90^89 = 8.46414978 Ã 10^173
So 89^90 is greater by a factor ~ 100
2006-10-11 08:35:02
·
answer #3
·
answered by Stuart T 3
·
0⤊
0⤋
I have proved that x^(1/x) decrease with x increasing for x >e
please see the link below
so 89^(1/89)> 90^(1/90)
raise them to power 89 * 90
89^90 > 90^89
2006-10-11 08:57:18
·
answer #4
·
answered by Mein Hoon Na 7
·
0⤊
0⤋
you can look at how it works with smaller numbers and see if there is a trend.
so look at
1^2 and 2^1
2^3 and 3^2
3^4 and 4^3
3^5 and 5^4
we get
1<2
8<9
81>64
1,024>625
15,625>7,776
you will see when you look at this that though the first 2 the secong formula is bigger as the numbers get bigger the 1st number is bigger than the second one.
This trend will follow hence 89^90 > 90^89
2006-10-11 08:39:32
·
answer #5
·
answered by Tom J 1
·
0⤊
0⤋
89^90 is 89*89*89... 90 times
while
90^89 is 90*90*90... 89 times... I guess they would be close as numbers go... so let's work it out
excel gives
2.7868E+175
8.4641E+173
so the first is greater..... 89^90 > 90^89
2006-10-11 08:37:06
·
answer #6
·
answered by blind_chameleon 5
·
0⤊
0⤋
The rote way is to just use the calculator
to find the values and see which is greater.
We will try something different.
Assume 89^90 is greater than 90^89
Taking logs we have
90 log89 gr 89 log 90
90/89 gr log90/log89
1.011 gr1.954/1.949
1.011gr1.002
What we assumed is true
Hence proved
2006-10-11 09:30:23
·
answer #7
·
answered by openpsychy 6
·
0⤊
0⤋
simple proof
89^90 >? 90^89 ... take log
90 log 89 >? 89 log 90
(90/89) >? (log 90 / log 89) .... this is true, thus proven
2006-10-11 08:38:24
·
answer #8
·
answered by cw 3
·
0⤊
0⤋
89^90=(89^2)^45
=8189^45
=(8189^44)*8189 ok.
now take 90^89=(90^88)*90=(8100^44)*90. now since 8189>8100and 8189>90 we can conclude that 89^90>90^89.
thanx maths kp. my answer sucks. 89 squared is 7921 and not 8189. so my proof is invalid. thinking of another solution.
2006-10-11 10:47:40
·
answer #9
·
answered by Anonymous
·
1⤊
0⤋