F4數學...

climbd321

1)It is given that 5^(x+1)=0.5.if5^(-kx)=8,find the value of k.
2)3^(x+3)-2(3^(x+2))=30-3^x
3)27(5^x)-125(3x)=0


4)logx^4/log9 ÷logx^5/log3 (x>0,y>0 andx<y=/=1

求解答

kingtom

1. 5^(x+1)=0.5
5^(x+1) = 5^-1
x+1 = -1
x=-2
5^(2k) = 8
2k log 5 = log 8
k = log8/2log5
督機數?

[ 本帖最後由 kingtom 於 10-1-2 02:46 PM 編輯 ]

kingtom

2.
3^(x+3)-2(3^(x+2))=30-3^x
3^(x+3)-2(3^(x+2))+3^x = 30
3^x(3^3-2(3^2)+1) = 30
3^x(10)=30
3^x=3
x=1

kingtom

3.
27(5^x)-125(3^x)=0
3^3*5^x - 5^3*3^x=0
3^3*5^x = 5^3*3^x
3^x/5^x = 5^3/3^3
(3/5)^x = (5/3)^3
             = (3/5)^-3
x = -3

kingtom

logx^4/log9 ÷logx^5/log3 (x>0,y>0 andx<y=/=1)
where is y?
同埋想求乜?

[ 本帖最後由 kingtom 於 10-1-2 02:46 PM 編輯 ]

climbd321

原帖由 kingtom 於 10-1-2 02:42 PM 發表
logx^4/log9 ÷logx^5/log3 (x>0,y>0 andx

simplify the following expressions

climbd321

原帖由 kingtom 於 10-1-2 02:37 PM 發表
1. 5^(x+1)=0.5
5^(x+1) = 5^-1
x+1 = -1
x=-2
5^(2k) = 8
2k log 5 = log 8
k = log8/2log5
督機數?

咁5^(x)=0.5.   
5^(-kx)=8又點計-3-

kingtom

等等幫你搞~

36661124

4)logx^4/log9 ÷logx^5/log3 (x>0,y>0 andx<y=/=1
=4logx/2log3  x log3/5logx
=2/5

climbd321

原帖由 36661124 於 10-1-2 05:35 PM 發表
4)logx^4/log9 ÷logx^5/log3 (x>0,y>0 andx

仲差1題系#6~麻煩哂-3-

kingtom

原帖由 climbd321 於 10-1-2 04:49 PM 發表

咁5^(x)=0.5.   
5^(-kx)=8又點計-3-

5^x = 0.5
x log 5 = log 1/2
x = (log1/2) / log5
5^{-k[(log1/2)/log5]} = 8
-k[(log1/2)/log5] log5 = log8
k = -log8/{log5[(log1/2)/log5]}
-.-怪怪的...

36661124

5^(x)=0.5
5^(-kx)=8

(5^x )^-k = 8
0.5^-k=8
(-k)(log0.5)=log8   <<<<<<<<<到ｅ步禁機就得
(-k)(log1 - log2)=3log2
-klog1  + klog2=3log2
                   k=3

[ 本帖最後由 36661124 於 10-1-2 06:20 PM 編輯 ]