2^a + a =3

首先，你可以用觀察的方法去找出來。但這當然是不能每次也做到。

另外就會有些人用些不知什麼方法，都是不對的。例如有人用2=log(100)，然後寫(log100)^a=a log 100，其實是錯誤的，log(100^a)=a log100才對。(log100)^a什麼也不能變，只能變作2^a。

其實要解這類的數，是不能靠普通處理數學的方法的。例如我將題目改成2^a + a = 4 你就不懂做了，而這題是有解的！解大約是1.3861669801。

所以，我們會用一種叫作數值法的方法處理這類問題。

我們會用其中一點開始，然後不斷向根迫近。
設f(x)=2^x + x -3
x(n+1)=x(n)-f(x(n))/f'(x(n))
這個的意思是在curve上找某一點上的切線然後計算這切線的x-intercept。

例如用上面的例子，
f'(x)=2^x ln2 + 1
經常大於0
代表這式最多有一個根。

用x(0)=0
用式子︰x(n+1)=x(n)-(2^x+x-3)/(2^x ln2+1)我們會分別找到
1.181232218
1.006674763
1.000008974
1.000000000
這時計數機已經不能再計下去了，因為我的計算機只有13個小數位。用這種方法就能找到式子的根。
這個數值法叫作牛頓法則，在工程學上有很大的用處。或者你可以用分半法 Bisection Method，但要做多很多才能找到根的所在。

當然，如果你用數值法找到1.000000000，你多數也會試試1是不是一個無限準確的答案吧。

Those answers in the above are flaw because some have problems in arithmetics and/or logic.

The question here is to solve 2^a + a = 3. In other words, we are going to find the roots of the equation 2^a + a - 3 = 0. But this is a non-linear function. Most non-linear function cannot be solved by an algebraic way (this question also cannot be solved algebraically). Therefore, some iterative methods are used to find an approximate answer to it, like Newton's Method. Hope these will help.

To solve an equation, we mean that we want to find all values of a so that the equation is satisfied.

First, note that 2^a is a strictly increasing function, meaning that the graph is always going upwards. Similarly, a - 3 is also a strictly increasing function. Therefore 2^a + a - 3 is also a strictly increasing function. Since a strictly increasing function is one whihc always goes upwards, there is only 1 root to the function.

By observation, we know that 2^1 + 1 - 3 = 0. Therefore, we know that 1 is a root of the equation. But since as above, we have shown that this equation has only 1 root, we know that 1 is the only solution to the question.

As a conclusion, there are many many different kinds of equations that cannot be solved by only calculation, like this question.

2^a + a = 3
2^a = 3 - a
a log 2 = log ( 3 - a )

then, a = 1 and log 2 = log ( 3 - a )
3 - a = 2
a = 1

咁咪兩次証明到a = 1囉

2006-11-25 09:49:48 補充：
faifai3120000抄我d字眼一模一樣ge.......

use log

frist) 2 = log100

log100^a + a =3
a x log100 + a = 3
a x 2 + a = 3
3a = 3
a = 1