M1求解（log+圖）~~20點

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 -^.^-

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！

~~~~~~~~~~~~~

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～

^___^