F.3 Laws of Indices-Exponential equations

a)
5 2^2y+1 - 3 2^2y+4 = -38912
5(2)^(2y+1)-3(2)^(2y+4) =-38912..........要知道2^(2y+1)=2(2)^(2y)
10(2)^(2y)-48(2)^(2y)=-38912
-38(2)^(2y)=-38912
(2)^(2y)=1024
4^y=1024
4^y=4^5
y=5

b) 2^y+1 = 2既y次方+1..............
2^y + 2^(y+1) + 2^(y+2) = 7/2
2^y + 2^(y+1) + 2^(y+2) = 7/2
2^y + 2(2^y) + 4(2^y) = 7/2
7(2^y) = 7/2
(2^y) = 1/2
(2^y) = 2^-1
y=-1

c)
[3^2y + 3^(y-1)] / [3^2y - 3^(y-1)] = -2
3^2y + 3^(y-1)= -2 [3^2y - 3^(y-1)]
3^2y + 3^(y-1)= -2(3^2y) + 2[3^(y-1)]
3(3^2y) = 3^(y-1)
3^(2y+1) = 3^(y-1)
log 3^(2y+1) = log 3^(y-1)
(2y+1) log 3 = (y-1) log 3
2y+1 =y-1
y=-2

注 : 有無學過log(對數) @@? 無學過話我知0.0

2008-11-03 16:46:20 補充：
log 全寫係 logarithms , 中文叫 對數
不過.....在實用性方面 如上一樣 可以用泥解方程0.0

log有好多性質, 不過....得一個性質對解方程有用
log a^n = n log a , a>0

用返c)泥廣
3^(2y+1) = 3^(y-1)
log 3^(2y+1) = log 3^(y-1)
(2y+1) log 3 = (y-1) log 3 .....用左log 睇返上面個性質 次方 會落左下面 所以會容易解

你9成9都吳會明架啦=_=| 都係用返d普通方法啦..........

3^(2y+1) = 3^(y-1)
2y+1=y-1
y=-2

5 x 2^2y+1 - 3 x 2^2y+4 = -38912
5*2^2y*2 - 3 * 2^2y*16 =-38912
2^2y(5*2-3*16)=-38912
2^2y(-42)=-38912
2^2y=1024
2^2y=2^10
2y=10
y=5

2^y + 2^y+1 + 2^y+2 = 7/2
2^y +2^y +2^y=1/2
3(2^y)=1/2
2^y=1/6
.......唔識



3^2y + 3^y-1 / 3^2y - 3^y-1 = -2
(3^y)(3^y)+3^y(-3)/(3^y)(3^y)-3^y(+3)=-2
(3^y)(3^y+1-3)/(3^y)(3^y-1+3)=-2
3^y-2/3^y+2=-2
3^y-2=-2(3^y+2)
y=-1