Two variables xand y are related by the equation y = ka^x ,where a and k are constants.
When x = loga（底）2,y= 8.If log2（底）y is plotted against x, the straight line obtained has slope equal to -1.
a,Find the values of a and k.(anw:a=1/2;k=4)
b,What is the intercept on the vertical axis of the line?(anw:2)
M1求解（log+圖）~~20點
2014-02-26 5:58 am
回答 (3)
2014-02-26 6:42 am
✔ 最佳答案
a)x=log(a)2 , y=8
y = ka^x
8 = k a^[log(a)2]
8 = 2k
k=4
y=4a^x
log y = log 4 + x log a
log(2)y = [log(2)a] x + log(2)4
Slope=log(2)a=-1
log(2)a=-1
log(2)a=log(2)(1/2)
a=1/2
b) x=0
log(2)y = [log(2)a] x + log(2)4
log(2)y = [log(2)a] (0) + 2
log(2)y = 2
∴The intercept is 2
2014-02-26 00:59:24 補充：
謝謝你們 -^.^-
2014-02-26 6:55 am
y = ka^x
log₂y = log₂k + log₂(a^x)
log₂y = log₂k + [log₂(a)]x
log₂y = [log₂(a)]x + log₂k
It is a form of y = mx + C
In a plot of log₂y against x, the slope is log₂(a), it is given slope = -1
==> log₂(a) = -1
==> a = (1/2) since 2^(-1) = 1/2
Given when x=loga2 ; y=8
loga2 = log(1/2)2 = -1 = x
sub into : log₂y = [log₂(a)]x + log₂k
==> log₂8 = (-1)(-1) + log₂k
==> 3 = 1 + log₂k
log₂k = 2
k = 4
intercept is log₂k = 2
2014-02-25 22:57:23 補充：
我也慢了，
明天刪吧！
~~~~~~~~~~~~~
log₂y = log₂k + log₂(a^x)
log₂y = log₂k + [log₂(a)]x
log₂y = [log₂(a)]x + log₂k
It is a form of y = mx + C
In a plot of log₂y against x, the slope is log₂(a), it is given slope = -1
==> log₂(a) = -1
==> a = (1/2) since 2^(-1) = 1/2
Given when x=loga2 ; y=8
loga2 = log(1/2)2 = -1 = x
sub into : log₂y = [log₂(a)]x + log₂k
==> log₂8 = (-1)(-1) + log₂k
==> 3 = 1 + log₂k
log₂k = 2
k = 4
intercept is log₂k = 2
2014-02-25 22:57:23 補充：
我也慢了，
明天刪吧！
~~~~~~~~~~~~~
2014-02-26 6:49 am
OK, 我慢了~
我刪帖, 請看以下:
https://s.yimg.com/rk/HA00430218/o/307265068.png
2014-02-26 16:50:35 補充：
HK~ 客氣了～
繼續努力作答，助人助己～
^___^
我刪帖, 請看以下:
https://s.yimg.com/rk/HA00430218/o/307265068.png
2014-02-26 16:50:35 補充：
HK~ 客氣了～
繼續努力作答，助人助己～
^___^
收錄日期: 2021-04-27 20:36:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140225000051KK00170