我有幾條關於laws of integral indices 嘅數學題唔識，
1a. 3^(n+1) over 3^(n-1)
2a. If 2^(x+y)=16, express x in terms of y.
2b.（聯立方程）Hence, if 上面果條係 2^(x+y)=16 ，下面果個係2^(x-y)=2，find the values of x and y.
希望能夠解答我以上問題，謝謝！
laws of integral indices 數學題
2010-10-01 11:32 pm
回答 (2)
2010-10-02 12:44 am
✔ 最佳答案
1a)[3^(n+1)] / 3^(n-1)= 3 ^ [(n+1) - (n-1)]= 3 ^ 2= 92a)2^(x+y) = 162^(x+y) = 2^4x + y = 4x = 4 - yb)2^(x+y)=16.......(1)2^(x-y)=2.......(2)由a) , (1) 中 可得 x = 4 - y , 代入 (2) :2^(4-y - y) = 22^(4 - 2y) = 2^14 - 2y = 1y = 3/2x = 4 - 3/2 = 5/2
2010-10-02 1:34 am
3^(n+1) over 3^(n-1)
3^n+1-n+1
3^2
9
______________
2^(x+y)=16
2^x+y=2^4
x+y=4
x=4-y
_______________
2^(x+y)=16------------1
2^(x-y)=2--------------2
From1:
x=4-y------3
Sub. 3 into 2:
2^4-2y=2^1
4-2y=1
y=1.5
so,x=2.5
3^n+1-n+1
3^2
9
______________
2^(x+y)=16
2^x+y=2^4
x+y=4
x=4-y
_______________
2^(x+y)=16------------1
2^(x-y)=2--------------2
From1:
x=4-y------3
Sub. 3 into 2:
2^4-2y=2^1
4-2y=1
y=1.5
so,x=2.5
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