Logarithmic Functions(F4Maths)

(a) log9*(3x)=log3*6
log 3x / 3log 3 = 3log6 / 3log3
log 3x = 3log6
x = 72

2.
Given that log2*3=p
so, log3/ log2 =p
log3 = p log2

(a)
log3*(2/3)
=log2-log3 / log 3
= log2 - p log2 / p log2
= 1-p / p

(b) log3*(1/12)
= log 1-log12 / log3
= 0 -( log3 + log2 + log2) / log3
= - (p+1+1) log 2 / p log 2
= -p-2 / p

(c) log3*(4.5)^-1
= log 2 - (log3 + log3) / log3
= 1-2p / p

2011-12-04 21:32:57 補充：
1 (a) cal wrong-.- s0r

correct answer is
log9*(3x)=log3*6
log 3x / 2log 3 = 2log 6 / 2log 3
log 3x = 2log6
x = 12

log(10) 100 = 2
log(a) b = c => b = a^c
let y = log(3)6 => 3^y = 6
take log(9) 3^y = log(9) 6
y log(9) 3 = log(9) 6 => y = log(9) 6 / log(9) 3
log(9) 3x= log(9) 6 / log(9) 3
[log(9) 3x] [log(9) 3] = log(9) 6
log(9) 9x= log(9) 6
9x = 6 => x = 2/3

same principle adopted in Q2

2011-12-04 16:38:03 補充：
log10 100 = 2
loga b = c => b = a^c
let y = log3 6 => 3^y = 6
take log9 3^y = log9 6
y log9 3 = log9 6 => y = log9 6 / log9 3
log9 3x = log9 6 / log9 3
[log9 3x] [log9 3] = log9 6
log9 9x = log9 6
9x = 6 => x = 2/3

log2 3 = p
log3 3 = p log3 2
1 = p log3 2
log3 2 = 1/p
a.
log3 [2/3]
log3 2 - log3 3 = 1/p - 1
b.
log3 [1/12]
-[log3 12] = -{log3 [2^2] + log3 [3]}
-{2log3 2 + 1} = -(2/p) - 1
c.
log3 [(4.5)^-1] = -log3 [4.5]
= -[log3 (9/2)]
= -[log3 (3^2) - log3 (2)]
= -[2log3 (3) - (1/p)]
= (1/p) - 2