F.4 math Log

Vicky

2009-11-24 4:58 am

小弟有幾題唔識做~請賜教

12. The fihure shows a graph of log y againsts log x for the relationship log=mlogx+c, where m and c are constants. the graph passes throught (2,2.22) and (5,6).
(a) Find the values of m and c.
(b)Express y in terms of x.
(c) Find the percentage change in the value of y when x is doubled.
圖片參考：http://imgcld.yimg.com/8/n/HA00367349/o/700911230128113873409890.jpg
30. Given that log底數4 2=x and log底數5 3 = y, express the following in terms of xand y.
(a) log底數2 3
(b) log底數27 72

31. Given that log底數25 10 =x, express log底數5 2 in terms of x.

圖片參考：http://imgcld.yimg.com/8/n/HA00367349/o/700911230128113873409901.jpg

圖片參考：http://f10.wretch.yimg.com/akpuser01/118/1598275930.jpg?nqS3gr5DcR857HU9Bui1gxgpVaNE6ILJICTV.dddlCLIA9nLY9vK8dBDlsmqWIw-

更新1:

to nelsonywm2000: 12b 的答案是 y = 0.501x^1.26 請問點計呢? thx~

回答 (1)

用戶9112

2009-11-24 5:53 am

✔ 最佳答案

12. (a) log y = m log x + c
2.22 = 2m + c ... (1)
6 = 5m + c ... (2)
(2) - (1) => 3.78 = 3m
m = 1.26
Sub into (1), 2.22 = 2.52 + c => c = -0.3
(b) log y = 1.26 log x - 0.3
log y = log x^(1.26) - log(1.995)
log y = log [x^(1.26) / 1.995]
y = x^(1.26) / 1.995
(c) Let original x, y be X and Y so that
Y = X^(1.26) / 1.995
new y = (2X)^(1.26) / 1.995 = [X^(1.26) / 1.995](2^1.26) = (2^1.26)Y
Percentage increase in y is (2^1.26Y - Y)/Y * 100% = 139.5%
(30) log(5) 2 means log 2 base 5
log(5) 2 = x => log2/log5 = x => log2 = xlog5
log(5) 3 = y => log3/log5 = y => log3 = ylog5
(a) log(2) 3
= log3/log2
= ylog5/(xlog5)
= y/x
(b) log(27) 72
= log72 / log27
= log(2^3*3^2)/log(3^3)
= [3log(2) + 2log(3)] / 3log(3)
= (3xlog5 + 2ylog5) /(3ylog5)
= (3x + 2y)/3y
= x/y + 2/3
(31) log(25) 10 = x
log10/log25 = x
log10 = xlog25
1 = 2xlog5
log5 = 1/2x
log(5) 2
= log2/log5
= log(10/5) / log5
= (log10 - log5) / log5
= 1/log5 - 1
= 2x - 1
(35) {[log(5) x^2] + [2log(1/5) √x]} / [log(5) √x]
= {[2logx / log5] + [log(1/5) x]} / [log√x / log5]
= {[2logx / log5] + [logx / log(1/5)]} / [(1/2)logx / log5]
= {[2logx / log5] - [logx / log5]} / [(1/2)logx / log5]
= {[2logx] - [logx]} / [(1/2)logx]
= 2
(41) [log(25) 2x] - [log(5) 8] = 0
log(25) 2x = log(5) 8
log(2x)/log25 = log8/log5
log(2x) / (2log5) = log8/log5
log(2x) = 2log8
log(2x) = log64
x = 32

2009-11-23 22:20:10 補充：
1/1.995 = 0.501!!!!

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