F4 maths 對數

若s = log 2 及t = log 3, 試以 s 和 t 表示下列各項。
1. log √ 72

log √72
= (1/2) log 72
= (1/2) log (8 x 9)
= (1/2) log (2^3 x 3^2)
= (1/2) (3 log 2 + 2 log 3)
= (1/2) (3s + 2t)


若 p = log 3及q = log 5, 試以 p 和 q 表示下列各項。
1. log 0.36
2. log√ 24
log 0.36
= log (36 / 100)
= log (9 / 25)
= log9 - log25
= log3^2 - log5^2
= 2log3 - 2log5
= 2p - 2q

log √24
= (1/2) log 24
= (1/2) log (3 x 8)
= (1/2) log (3 x 1000 / 125)
= (1/2) [log 3 - log 125 + log 1000]
= (1/2) [log 3 - log 5^3 + log 10^3]
= (1/2) [log 3 - 3 log 5 + 3 log 10]
= (1/2) (p - 3q + 3)

其實尾二那步還沒有錯的...

螞蟻雄兵 ( 知識長 )

In question 2

the answer should be (p-3q+3)/2 !

若s=log2 及t=log3，試以s和t表示log√72
log√ 72
=(1/2)log72
=(1/2)log[(2^3)*(3^2)]
=(3/2)log2+log3
=(3/2)s+t

若 p=log3及q=log5，試以p和q表示下列各項
1. log 0.36
Sol
q=log5=1－log2
log2=1－q
log0.36
=log36－log100
=log[(2^2)*(3^2)]－2
=2log2+2log3－2
=2(1－q)+2p－2
=2p－2q

2. log√ 24
Sol
log√ 24
=(1/2)log24
=(1/2)log[(2^3)*3]
=(3/2)log2+(1/2)log3
=(3/2)*(1－q)+(1/2)p
=p/2－q/2+3/2