小弟唔係幾識做以下幾題數,希望有心人可以幫幫我.. 先謝

記住log(x^a)=a log x
當你轉晒做log x個時,你就會計到

brianlhm :記住log(x^a)=a log x
當你轉晒做log x個時,你就會計到

我知,但係我最後都係計唔到一個整數
第一條我計到-5/6 log x...
第二條我計到6n+3...
但係兩題既答案係3/8 and 9

1.

JN568@80K :

brianlhm :記住log(x^a)=a log x
當你轉晒做log x個時,你就會計到

我知,但係我最後都係計唔到一個整數
第一條我計到-5/6 log x...
第二條我計到6n+3...
但係兩題既答案係3/8 and 9

不如你post你個計法出黎
睇下點解會計錯

JY4066 :1.

唔該哂版友
BTW你用咩software黎打?好靚~

2.
=[ (n-1)log x + (2-n) log x ]/ (1/3) log x
= 3 log x / (1/3) log x
=9

I don't think the answer of Q.2 is 9, it should be 3.

Rafael Chicharito :2.
=[ (n-1)log x + (2-n) log x ]/ (1/3) log x
= [del]3[/del] log x / (1/3) log x
=[del]9[/del]3

JN568@80K :唔該哂版友
BTW你用咩software黎打?好靚~

其實我係用Excel打 ...

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第2題我同樓上一樣都係計到3，所以之前冇貼。

Utopia@TMS~ :I don't think the answer of Q.2 is 9, it should be 3.

Rafael Chicharito :2.
=[ (n-1)log x + (2-n) log x ]/ (1/3) log x
= [del]3[/del] log x / (1/3) log x
=[del]9[/del]3

師兄想問下(n-1)log x+(2-n)log x點變?
小弟唔係幾明..
thx

JN568@80K :

Utopia@TMS~ :I don't think the answer of Q.2 is 9, it should be 3.

Rafael Chicharito :2.
=[ (n-1)log x + (2-n) log x ]/ (1/3) log x
= [del]3[/del] log x / (1/3) log x
=[del]9[/del]3

師兄想問下(n-1)log x+(2-n)log x點變?
小弟唔係幾明..
thx

舉個例：
　2 log x + 3 log x
= 5 log x

所以：
　(n-1)log x+(2-n)log x
= [(n-1)+(2-n)] log x
= (n-1+2-n) log x
= log x

JN568@80K :師兄想問下(n-1)log x+(2-n)log x點變?

Do you mean you are not sure how logx^(n-1)+logx^(2-n) can change to (n-1)log x+(2-n)log x?

Remember the rule of log: logm^n=nlogm will do. This rule is very useful.

Then you follow the steps in #11 and solve the problem.

我明點做啦~
唔該哂樓上兩位師兄幫忙