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 :-)
 
Last edited by a moderator:

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
: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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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:
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
5,450
21
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)...
Well, the posts that we make aren't going anywhere. You can post your work during (your) daytime when you're fresh, and then I'll see it during (my) daytime.

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
No. I don't know why you say "the y inercept would be z-2". There's no z here! But you've almost got the number right, you just forgot to include the log. You should have gotten:

[tex]b=\log_3\left(\frac{5}{9}\right)-2[/tex]
 

Want to reply to this thread?

"Log in a diagram" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top