Basic logarithm Help Please!
Favorites|Homepage
Subscriptions | sitemap
HOME > > Basic logarithm Help Please!

Basic logarithm Help Please!

[From: ] [author: ] [Date: 11-12-06] [Hit: ]
-For 9, answer is 5.......
2. Fill in the blank: ln e3 = _____.







9. Solve the following equation for x:

logx +log(x-1) = log(4x)

(Points : 2)
x = 5

x = 0,5 <----- I think it is this one!

x = 1/4log(x - 1)

There are no solutions.


PLEASE UR WORK?

-
ln e3 = _____

I suppose you mean ln e^3 =

As you know, if log_b (m) = a, then this means that b^a = m, in which "b" is the base of the logarithm. In your example, ln means that the base is "e", so it's like writing log_e (e^3), and therefore the answer is simply 3 because of the property mentioned above.

logx +log(x-1) = log(4x)

log [x(x - 1)] = log(4x)

x(x - 1) = 4x -----> x^2 - x = 4x -----> x^2 - 5x = 0 ----->

x(x - 5) = 0

x1 = 0
x2 = 5

You must ignore the answer x = 0 because when you put it into the original equation you get logarithms of zero or negative numbers, which are undefined. So your answer is x = 5

You see, with x = 5 this is what you get in the original equation:

log 5 + log (5 - 1) = log (4*5)

log 5 + log 4 = log(4*5)

log (5*4) = log (4*5)

log 20 = log 20

So that confirms the answer...

-
For 9, answer is 5.
1
keywords: Basic,Help,logarithm,Please,Basic logarithm Help Please!
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .