Log in a diagram

1. The problem statement, all variables and given/known data

http://img21.imageshack.us/img21/2327/nummer1.jpg [Broken]

2. Relevant equations

6(a) y=2 so the answer should about 3,3

3. The attempt at a solution
6(b)
Proplem is that i dont know what the opposite(invert) of log3 is...
then i could take that opposite and use it on y... a litle help here please :-)
 
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Tom Mattson

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Proplem is that i dont know what the opposite(invert) of log3 is...
The inverse of a logarithmic function is an exponential function. That is, let [itex]a>0[/itex], [itex]a\neq 1[/itex]. Then if [itex]f(x)=log_a(x)[/itex] then [itex]f^{-1}(x)=a^x[/itex].

Does that help?
 
no (sorry), i think i was wrong, and i am utterly and completly lost... would apreciate a litle help to get back on track...
 

Tom Mattson

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OK, try this.

1.) Write down [itex]y[/itex] as a function of [itex]x[/itex]. You should be able to do this using part (b) of the given information. Call this function [itex]f(x)[/itex] for now.

2.) Note that [itex]y=\log_3(z)[/itex], as stated in the problem.

Since [itex]y=f(x)[/itex] and [itex]y=\log_3(z)[/itex] what can you conclude about [itex]z[/itex] as a function of [itex]x[/itex]?
 
well, if y is a function of z then f(x) must also be function of z

so, f-1(x)--> z=log3(y)? hmm...
 
wait, f(z)=x?
 

Tom Mattson

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No, [itex]y=f(x)[/itex]. [itex]z[/itex] is some other function of [itex]x[/itex] (call it [itex]g(x)[/itex]. But first thing's first: You need to find [itex]y=f(x)[/itex]. It should be easy--its graph is a straight line!
 
first i have to remove the log, that is to put it into y side of the equation. agh, sorry. It's the midle of the night were i am sitting rigth now, so i am strugeling a bit.
 

Tom Mattson

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:confused:

Forget the log for a minute. I'm asking you to find [itex]y[/itex] as a function of [itex]x[/itex]. They gave you the slope of the line and a point that the line passes through. Do you know how to write down the equation of a line?
 
what is a gradient? (it is the midle of the nigth here...)
 

Tom Mattson

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gradient=slope
 
errr, y=ax+b? and, ohh, i remember gradient is the slope of line wich is a. so in this case y=2x+b?
 
then 5/9=2x+b?
 

Tom Mattson

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fot that point, that is...
 
so i have to make quess on what x is in this given case? wich would be 2,3...
 
wich would make b=-4.04444...
 
to sum up; the answer should be z=2*2.3-4.0444...
Is this wrong? somebdy please check this answer! :-)
 

Tom Mattson

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hostergrad, if you don't know how to write down the equation of a line given the slope and a point then this whole enterprise is pretty hopeless.

fot that point, that is...
No, for any point on the line. A slope and a point uniquely determines a straight line.

so i have to make quess on what x is in this given case? wich would be 2,3...
No, they gave you the coordinates of a point on the line:

[tex](x,y)=\left(1,log_3\left(\frac{5}{9}\right)\right)[/tex]

wich would make b=-4.04444...
No, and I have no idea of how you arrived at that.

to sum up; the answer should be z=2*2.3-4.0444...
Is this wrong? somebdy please check this answer! :-)
It's wrong. We can continue once you give me the equation of that line. I'm not going to respond to any more random guessing.
 
ohh, wait. I wass up 2'o clock in hte night doing this and i somehow tough 1=y. really, staying up that late melts my brain... but what can i do? I live in eroupe and that's the only time the american conticent are awake (aviable at the computer, that is)...
ok so we get the equation:

y=2(1)+b

then; b is the y intercept.

so the y inercept would be z-2 (the 2 is the slope) or (5/9)-2?
that would mean that b=-13/9
wich gives us that:

z=2x-(13/9)=5/9

but that being said, im still not sure that i found b the rigth way. So migth not correct either...
Sorry for being stupid:frown: , but at least i am trying to understand it... tougth i can understand that it is frustrating when i seem completly unable to grasb this. Yes, my ability to remember math formula is a bit bad... thats why i'm here, because i dont have the people skils to get classmates to help me...:confused: but I do really apreciate the help!:approve:
 

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