ce Maths
2009-01-03 4:51 am
Solve log (base 2) (2x+3)+ log (base 2) (x-2)=2
Simplify log (a^2*b)/[log a + log √b), where a,b >1
{(4√2^x) * [4^(y/4)]} / [8^(z/2)]=
回答 (2)
2009-01-03 9:31 am
✔ 最佳答案
The answer is as follows:圖片參考：http://i320.photobucket.com/albums/nn347/old-master/090103_mat_1.jpg?t=1230917456
2009-01-03 5:55 am
a) Solve log (base 2) (2x+3)+ log (base 2) (x-2)=2
log2 [(2x+3)(x-2)] = 2
(2x+3)(x-2) = 4
2x2 + 3x - 4x - 6 - 4 = 0
2x2 - x - 10 = 0
(2x-5)(x+2) = 0
x = 5/2 or
x = -2 (not applicable)
圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_18.gif
b) Simplify log (a^2*b)/[log a + log √b), where a,b >1
log (a^2*b)
= log[a*√b]^2
= 2 log [a*√b]
= 2 [log a + log √b]
So, log (a^2*b)/[log a + log √b) = 2
圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_11.gif
c) {(4√2^x) * [4^(y/4)]} / [8^(z/2)]
= ???
2009-01-02 21:55:43 補充：
Sorry that the base 2 is a bit confusing.....
log2 [(2x+3)(x-2)] = 2
(2x+3)(x-2) = 4
2x2 + 3x - 4x - 6 - 4 = 0
2x2 - x - 10 = 0
(2x-5)(x+2) = 0
x = 5/2 or
x = -2 (not applicable)
圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_18.gif
b) Simplify log (a^2*b)/[log a + log √b), where a,b >1
log (a^2*b)
= log[a*√b]^2
= 2 log [a*√b]
= 2 [log a + log √b]
So, log (a^2*b)/[log a + log √b) = 2
圖片參考：http://tw.yimg.com/i/tw/ugc/rte/smiley_11.gif
c) {(4√2^x) * [4^(y/4)]} / [8^(z/2)]
= ???
2009-01-02 21:55:43 補充：
Sorry that the base 2 is a bit confusing.....
收錄日期: 2021-04-29 18:14:46
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https://hk.answers.yahoo.com/question/index?qid=20090102000051KK01908