F.4 log questions(20分)

[匿名]

2009-11-07 10:36 am

1. (a) Consider the function y= ka^x where a is larger than 0. When x=2, y=25; when x=4, y=6.25.

(b) Hence find the value of y when x=8.

2. (a) If log(base 4) x = y for x is not equal to 1 and x is larger than 0, express x in terms of y.
(b) By using the result of (a), find the value of log(base x)4 and express the answer in terms of y.
(c) Solve the equation log(base 4) x + log(base x)4 = 13/6 . (Correct your answers to 3 sig. fig. if necessary.)

3. A company plans to sell a new brand of MP3 player. After a t-week promotion, the rate P of youngsters recognizing this brand is given by P= 1- 5^(-0.2t).
(a) Find the rate of youngsters recognizing this brand after each of the following promotion period.
(i) 1 week (ii) 2 weeks (iii) 4 weeks
(b) If this company expects this brand to be recognized by 90% of youngsters, what should be the duration of the promotion.
(c) As the value of t increases, what are the characteristics of the values of P?
(Correct your answers to 1 d.p. if necessary.)

回答 (1)

用戶10012

2009-11-07 3:42 pm

✔ 最佳答案

1.
25 = ka^2...........(1)
6.25 = ka^4..........(2)
(2)/(1) 25/6.25 = a^2 = 4 so a = 2 ( a > 0)
Sub into (1) 25 = k2^2 = 4k, so k = 25/4
that is y = 25(2^x)/4
when x = 8, y = 25(2^8)/4 = 1600.
2.
(a) log(base 4)x = y, that is 4^y = x
(b) Let log(base x)4 = z, that is x^z = 4, so (4^y)^z = 4,
4^(yz) = 4,
zy = 1, z = 1/y.
(c)
log(base4)x + log(base x)4 = 13/6
so y + 1/y = 13/6
6y^2 + 6 = 13y
6y^2 - 13y + 6 = 0
(3y - 2)(2y - 3) = 0
y = 2/3 or 3/2
so x = 4^(2/3) = 2.52
or x = 4^(3/2) = 8
3
a1) P = 1 - 5^(- 0.2 x 1) = 1 - 5^(- 0.2) = 1 - 0.725 = 0.275.
a2) P = 1 - 5^(-0.2 x 2) = 1 - 5^(-.0.4) = 1 - 0.525 = 0.475
a3) P = 1 - 5^(-0.2 x 4) = 1 - 5^(-0.8) = 1 - 0.276 = 0.724
b) 0.9 = 1 - 5^(-0.2t)
0.1 = 5^(-0.2t)
-0.2t = log0.1/log5 = - 1.4307
so t = 1.4307/0.2 = 7.15 weeks.
c) characteristic values ?

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