指數ge問題...

[10^(1/3)]^y=2=10^(y/3) [ (a^b)^c=a^(bc) ]
(a) 32/10^(2y) = 32/{[10^(y/3)]^6} = 32/(2^6)=32/64=1/2 (or 0.5)
(b) 100^(-y)=(10^2)^(-y)=10^(-2y)=[10^(y/3)]^(-6)=2^(-6)=1/64

2008-01-05 23:35:27 補充：
To natcny:第1題(a)10^y=8及10^(2y)=64雖然都啱,但人地唔係問呢樣野;第2題,10*10^(-y) =10*8(-1)唔啱,第一步應該係(10*10)^(-y)=[10^(-y)]^2

[10^(1/3)]^y=2
10^(y/3)=2
10^y=2^3
10^y=8
(a) 10^(2y)=8^2
10^(2y)=64
(b) 10* 10^(-y) = 10* 8(-1)
100^(-y) = 10/8
100^(-y) = 1.25