請問這條題目怎麼計...

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http://i.imgur.com/iycQX.gif

希望我能理解你的題目吧 一堆可能看不懂
Let u = 2^x , then we have
u^2 + 7u - 3 = 0
u = (-7 +- root(7^2 - 4(1)(-3)))/2(1)
u = (-7 +- root61)/2
Hence,
2^x = (-7 + root61)/2 or
2^x = (-7 - root61)/2 (reject)
beacuse (-7 - root61)/2 <0, log (-7 - root61)/2 is undefined
x log 2 = log (-7 + root61)/2
x = log ((-7 + root61)/2) / log 2
x = -1.30

2011-04-25 04:33:39 補充：
上面的是理解成
2^(2x) +1 + 7(2^x) - 4 = 0

以下的是理解成
2^(2x+1) + 7(2^x) -4 =0

Let u = 2^x , then we have
2(u^2) + 7u - 4 = 0
(2u - 1)(u + 4) = 0
u = 1/2 or u = -4 (reject) ||| log -4 is undefined
2^x = 1/2
x log 2 = - log 2
x = -1

2011-04-25 05:14:06 補充：
以上的我看成2^(2x) + 1 + 7(2^x) - 4 = 0
以下的我看成2^(2x+1) + 7(2^x) - 4 = 0
Let u = 2^x , then we have
2u^2 + 7u - 4 = 0
(2u - 1)(u + 4) = 0
u = 1/2 or u = -4(reject) ||| log -4 is undefined

2^x = 1/2
x log 2 = - log 2
x = -1

2011-04-25 05:14:56 補充：
不好意思= =不知道為甚麼刷新後沒顯示 所以多打了一次~