#### hostergaard

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

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

2. Relevant equations

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 $a>0$, $a\neq 1$. Then if $f(x)=log_a(x)$ then $f^{-1}(x)=a^x$.

Does that help?

#### hostergaard

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
Gold Member
OK, try this.

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

2.) Note that $y=\log_3(z)$, as stated in the problem.

Since $y=f(x)$ and $y=\log_3(z)$ what can you conclude about $z$ as a function of $x$?

#### hostergaard

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
Gold Member
No, $y=f(x)$. $z$ is some other function of $x$ (call it $g(x)$. But first thing's first: You need to find $y=f(x)$. It should be easy--its graph is a straight line!

#### hostergaard

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
Gold Member

Forget the log for a minute. I'm asking you to find $y$ as a function of $x$. 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?

#### hostergaard

what is a gradient? (it is the midle of the nigth here...)

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#### hostergaard

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
Gold Member
errr, y=ax+b? and, ohh, i remember gradient is the slope of line wich is a. so in this case y=2x+b?
Yes.

then 5/9=2x+b?
No. You need to use the given point on the line to find b.

#### hostergaard

fot that point, that is...

#### hostergaard

so i have to make quess on what x is in this given case? wich would be 2,3...

#### hostergaard

wich would make b=-4.04444...

#### hostergaard

to sum up; the answer should be z=2*2.3-4.0444...

#### Tom Mattson

Staff Emeritus
Gold Member
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:

$$(x,y)=\left(1,log_3\left(\frac{5}{9}\right)\right)$$

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...
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.

#### hostergaard

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 , 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... but I do really apreciate the help!

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