Question

指数函数

2^a×3^b=2^c×3^d=6，求证：（a-1）×（d-1）=（c-1）×（b-1）

Answer 1

同时除以6
2^a/2×3^b/3=2^c/2×3^d/3=1
2^(a-1)×3^(b-1)=2^(c-1)×3^(d-1)=1
取对数
(a-1)lg2+(b-1)lg3=(c-1)lg2+(d-1)lg3=0
所以
(a-1)lg2=-(b-1)lg3
(c-1)lg2=-(d-1)lg3
相除
(a-1)/(c-1)=(b-1)/(d-1)
所以(a-1)(d-1)=(c-1)(b-1)

Answer 2

=5^1÷5^log0.2(3)
=5÷0.2^-log0.2(3)
=5÷0.2^log0.2(3^-1)
=5÷0.2^log0.2(1/3)
=5÷1/3
=15

换底公式
=(lg3/lg4)(lg2/lg9)-lg32^(1/4)/lg(1/2)
=(lg3/2lg2)(lg2/2lg3)-lg2^(5/4)/(-lg2)
=(1/2*1/2)(lg3/lg2)(lg2/lg3)+(5/4)lg2/lg2
=1/4+5/4
=3/2