1. If log2 = a and log9 = b, express each of the following in terms of a and b.
a) log 3/2
b) log 54
c) log 0.12
2. Given that 8^x = 0.08^y = 1000. Without finding out the values of x and y, find the value of 1/x - 1/y.
3. Factorize the following: x^3 + x - 2/x - 8/(x^3)
4. IF log x : log y = m:n, then x = ?
maths F.4
2007-08-23 5:36 pm
回答 (2)
2007-08-23 6:30 pm
✔ 最佳答案
(1)圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug07/Crazyeqn16.jpg
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參考: My Maths knowledge
2007-08-23 7:06 pm
1 log 2 =a
log 9 = b
2log 3= b
log 3 = b/2
a) log 3/2 = log 3 - log 2 =b/2-a
b) log 54 = log (3x3x3x2) = log 3 + log 3 + log 3 + log 2 = 3b/2 + a = 1.5b + a
c) log 0.12 = log (3x2x2)/100 = log 3 + log 2 + log 2 - log 100 = b/2 + 2a - 2
2. 8^x = 1000
xlog8=log1000
xlog8 = 3
x = 3/log8--- (1)
0.08^y = 1000
ylog0.08 = log1000
y(log8 - log100) = 3
y=3/(log 8 - 2) --- (2)
Apply (1) and (2) into:
1/x - 1/y = 1/ (3/log 8) - 1/(3/log 8 - 2)
= log8 / 3 - (log 8 - 2) /3
= 2/3
3. x^3 + x - 2/x -8/(x^3)
= x^3 - 8/(x^3) + x - 2/x
= (x - 2/x) ([x^2 +x*2/x + (2/x)^2] +x - 2/x
= (x-2/x)[x^2+2+(2/x)^2] + (x-2/x)
=(x-2/x)[x^2 +3 +(2/x)^2]
4. logx/logy=m/n
nlogx=mlogy
logx^n = logy^m
x^n=y^m
x=y^(m/n)
log 9 = b
2log 3= b
log 3 = b/2
a) log 3/2 = log 3 - log 2 =b/2-a
b) log 54 = log (3x3x3x2) = log 3 + log 3 + log 3 + log 2 = 3b/2 + a = 1.5b + a
c) log 0.12 = log (3x2x2)/100 = log 3 + log 2 + log 2 - log 100 = b/2 + 2a - 2
2. 8^x = 1000
xlog8=log1000
xlog8 = 3
x = 3/log8--- (1)
0.08^y = 1000
ylog0.08 = log1000
y(log8 - log100) = 3
y=3/(log 8 - 2) --- (2)
Apply (1) and (2) into:
1/x - 1/y = 1/ (3/log 8) - 1/(3/log 8 - 2)
= log8 / 3 - (log 8 - 2) /3
= 2/3
3. x^3 + x - 2/x -8/(x^3)
= x^3 - 8/(x^3) + x - 2/x
= (x - 2/x) ([x^2 +x*2/x + (2/x)^2] +x - 2/x
= (x-2/x)[x^2+2+(2/x)^2] + (x-2/x)
=(x-2/x)[x^2 +3 +(2/x)^2]
4. logx/logy=m/n
nlogx=mlogy
logx^n = logy^m
x^n=y^m
x=y^(m/n)
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