Question

数学指数的运算

Answer 1

1.已知2^a*5^b=10
两边以10为底去对数,设lg2=t,则lg5=1-t
alg2+blg5=1
at+b(1-t)=1
(a-1)t+t+(b-1)(1-t)+(1-t)=1
(a-1)t+(b-1)(1-t)=0
(a-1)/(b-1)=1-1/t
2^c*5^d=10同理
(c-1)/(d-1)=1-1/t
所以(a-1)/(b-1)=(c-1)/(d-1)
(a-1)(d-1)=(b-1）(c-1)
2.f(x)=(a^x-a^-x)/(a^x+a^-x)=(a^2x-1)/(a^2x+1)
f(x)+f(y)=(a^2x-1)/(a^2x+1)+(a^2y-1)/(a^2y+1)
=[a^2(x+y)+a^2x-a^2y-1+a^2(x+y)-a^2x+a^2y-1]/[(a^2x+1)(a^2y+1)]
=2[a^2(x+y)-1]/[(a^2x+1)(a^2y+1)]
f(x)f(y)=(a^2x-1)/(a^2x+1)*(a^2y-1)/(a^2y+1)
=[a^2(x+y)-a^2x-a^2y+1]/[(a^2x+1)(a^2y+1)]
f(x)f(y)+1=[a^2(x+y)-a^2x-a^2y+1]/[(a^2x+1)(a^2y+1)]+1
=[a^2(x+y)-a^2x-a^2y+1+(a^2x+1)(a^2y+1)]/[(a^2x+1)(a^2y+1)]
=2[a^2(x+y)+1]/[(a^2x+1)(a^2y+1)]
[f(x)+f(y)]/[f(x)f(y)+1]=[a^2(x+y)-1]/[a^2(x+y)+1]=f(x+y)
所以f(x+y)=[f(x)+f(y)]/[f(x)f(y)+1]

Answer 2

经常会遇见把低次幂向高次幂转化的运算技巧。比如9的12次方转化为3的24次方这种运算。在判断能否被某个数整除的时候经常用到。
这个问题数学指数幂的运算,好难啊，辛辛苦苦回答了，给我个满意答案把