Login Diagram Homework: Solving Logarithmic Equations

[hostergaard]

Homework Statement

http://img21.imageshack.us/img21/2327/nummer1.jpg

Homework Equations

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

The Attempt at a Solution

6(b)
Proplem is that i don't know what the opposite(invert) of log₃ is...
then i could take that opposite and use it on y... a litle help here please :-)

 

[quantumdude]

Proplem is that i don't 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 completely lost... would apreciate a litle help to get back on track...

 

[quantumdude]

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...

 

[hostergaard]

wait, f(z)=x?

 

[quantumdude]

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.

 

[quantumdude]

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...)

 

[quantumdude]

gradient=slope

 

[hostergaard]

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

 

[hostergaard]

then 5/9=2x+b?

 

[quantumdude]

errr, y=ax+b? and, ohh, i remember gradient is the slope of line which 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? which 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...
Is this wrong? somebdy please check this answer! :-)

 

[quantumdude]

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? which 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...
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.

 

[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, I am 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 completely unable to grasb this. Yes, my ability to remember math formula is a bit bad... that's why I'm here, because i don't have the people skils to get classmates to help me... but I do really apreciate the help!

 

[quantumdude]

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:

$$b=\log_3\left(\frac{5}{9}\right)-2$$