logarithmic functions


by Erin_Sharpe
Tags: functions, logarithmic
Erin_Sharpe
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#1
Jan26-05, 04:23 PM
P: 17
y = logx 4

identify the domain and range

state whether the function is increasing or decreasing, and identify the vertical asympotote, teh x-intercept and the value of x at y = 1

I need some help!

thank you in advance!
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mathman
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#2
Jan26-05, 04:27 PM
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The 4 in your expression appears to be hanging out in space. It would be helpful if you could write the equation more clearly.
dextercioby
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#3
Jan26-05, 04:28 PM
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So your function is:
[tex] y(x)=\log_{x} 4 [/tex]

Can you determine the domain and range...??

Discuss the sign of its derivative...


Daniel.

P.S.Can u solve the equation y(x*)=1??Find x*...

HallsofIvy
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#4
Jan26-05, 06:15 PM
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Thanks
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logarithmic functions


I believe that loga(x) is only defined for a> 0. Does that help.
Zurtex
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#5
Jan26-05, 06:58 PM
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Also I don't think that loga(x) is defined for a=1.
dextercioby
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#6
Jan27-05, 03:54 AM
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Write
[tex] y(x)=\frac{\ln 4}{\ln x} [/tex]

Cau u see now what is the range??What about the domain?

Daniel.
Erin_Sharpe
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#7
Jan27-05, 07:14 AM
P: 17
See I thought x was 2??
dextercioby
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#8
Jan27-05, 07:21 AM
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What "2"??Where did u pick it out???You mean the solution to the equation
[tex] \frac{\ln 4}{\ln x} =1 [/tex]

I hope you're only kidding...

Daniel.
Erin_Sharpe
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#9
Jan27-05, 11:20 AM
P: 17
no, i'm kidding. this problem might as well be in chinese for all i understand of it
Erin_Sharpe
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#10
Jan27-05, 11:21 AM
P: 17
*** NOT kidding
dextercioby
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#11
Jan27-05, 11:27 AM
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Write it like that
[tex]\ln 4=\ln x [/tex]

Use the fact that logaritm is a surjective function...

Daniel.
Erin_Sharpe
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#12
Jan27-05, 11:28 AM
P: 17
i'm sorry i'm not trying to be difficult i just completely dont understand what you are talking about.
dextercioby
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#13
Jan27-05, 11:36 AM
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Quote Quote by Erin_Sharpe
i'm sorry i'm not trying to be difficult i just completely dont understand what you are talking about.
Okay.Do you know the relation between the exponential and the natural logarithm???If so,apply exponential on both sides of the last equation and tell me what u get.

Daniel.


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