Friday, June 8, 2007
Haha for the first time in a 1000 years, it's 2.20 am and I am actually NOT emo. But too bad, there's only junyao online to talk to and he wanna make me emo. What the hell.
And for some reason, I suddenly feel like talking about Maths on my blog.
My ex-classmate was asking me this question.
3^x + 4^x = 5^x
Looks so easy, all have same power before I realised I couldn't solve it!
From Pythagoras and Fermat's Last Theorem, the answer can be deduced as 2 easily. So now I shall dig my brain and look for the answer assuming Pythagoras and Last Theorem cannot be applied here.
3^x+4^x= 5^x
log(3^x+4^x)=log(5^x) = x log(5)
x=log(3^x+4^x)/log(5)=f(x)
guess x=1.9
1.9 1.920359 1.936563 1.949464 1.959738 1.967921 1.974439 1.979631
1.983768 1.987065 1.989691 1.991785 1.993453 1.994782 1.995841 1.996685
1.997358 1.997894 1.998322 1.998662 1.998934 1.99915 1.999323 1.99946
1.99957 1.999657 1.999727 1.999782 1.999826 1.999861 1.999889 1.999912
1.99993 1.999944 1.999955 1.999964 1.999971 1.999977 1.999982 1.999985
1.999988 1.99999 1.999992 1.999994 1.999995 1.999996 1.999997 1.999997
1.999998 1.999998 1.999998 1.999999 1.999999 1.999999 1.999999 1.999999
guess x=2.1
2.1 2.079759 2.063606 2.050718 2.040438 2.032239 2.025701 2.020487
2.016331 2.013017 2.010376 2.00827 2.006592 2.005254 2.004187 2.003337
2.00266 2.00212 2.001689 2.001346 2.001073 2.000855 2.000681 2.000543
2.000433 2.000345 2.000275 2.000219 2.000174 2.000139 2.000111 2.000088
2.00007 2.000056 2.000044 2.000035 2.000028 2.000022 2.000018 2.000014
2.000011 2.000009 2.000007 2.000005 2.000004 2.000003 2.000003 2.000002
2.000001 2.000001 2.000001 2 2 2 2 *
Approximate/guess 3^x+4^x=5^x to 2*[(3+4)/2]^x=5^x
2*[(3+4)/2]^x=5^x
2=[2*5/(3+4)]^x=[10/7]^x
x=log2/log(10/7)=log2/[log10-log7]
x=1.943358210 = 2
Woohoo. There's your answer Lionel. And no, I don't know what I am typing as well. =.=
Emo-ed
2:17 AM
