F.4 Law of Indics

2009-12-09 12:14 am
Solve:

1.3^2x+1 - 3^x - 4 = 0

2. 3x (3^2x) - 28x (3^x) + 9 = 0

回答 (1)

2009-12-09 12:19 am
✔ 最佳答案
1. 3^(2x + 1) - 3^x - 4 = 0

3(3^2x) - 3^x - 4 = 0

3[(3^x)^2] - 3^x - 4 = 0

Set y = 3^x

Then

3y^2 - y - 4 = 0, it is a quadratic equation,

(3y - 4)(y + 1) = 0

y = 4/3 or -1

3^x = 4/3 or 3^x = -1 (rejected)

3^x = 4/3

log(3^x) = log(4/3)

xlog3 = log(4/3)

x = log(4/3) / log3


2. 3(3^2x) - 28(3^x) + 9 = 0

3(3^x)^2 - 28(3^x) + 9 = 0

which is a quadratic equation again

(3(3^x) - 1)(3^x - 9) = 0

3^x = 1/3 or 3^x = 9

3^x = 3^(-1) or 3^x = 3^2

x = -1 or 2
參考: Physics king


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