Solving Logarithmic Equations Involving x & y

[Twisted--]

Hi :)
This is my first time using this forum, so thanks for your help in advance.

Homework Statement

Use the laws of logs to write y as a function of x for each. Then state domain.

a) log(xy) = 2 log(x-3)
b) log(y) + 3 = log(y + 1) + log(x)
c) log (x²/y) = 2 log(x + 5)

Answers:
a) y = (x-3)² / x
b) y = x / (1000 -x)
c) y = x² / (x + 5)²

The Attempt at a Solution

I got the first one by:

log(xy) = 2 log(x-3)
logx + log y = 2 log(x-3)
log y = 2 log(x-3) - logx
log y = log [ (x-3)² / x]

I'm not sure about is how they got rid of the "log" to get only x and y in the answer, though. Are you even allowed to divide "log" and cross it out?
I don't know where they got 1000 from in b).
I started by moving all the logs that contain "y" to the left side, but I didn't know where to go from there.
And I'm not sure how to do c).

 

[symbolipoint]

[strike]Recheck the original equation for item b in case you miscopied or misread.[/strike]

Item #c:

On right hand side, \[<br /> 2\log (x + 5) = \log (x + 5)^2 <br /> \]<br />

Continuing this from original equation,
\[<br /> \log \left( {\frac{{x^2 }}{y}} \right) = \log (x + 5)^2 <br /> \]<br />

therefore, \[<br /> \begin{array}{l}<br /> \frac{{x^2 }}{y} = (x + 5)^2 \\ <br /> x^2 = y(x + 5)^2 \\ <br /> y = \frac{{x^2 }}{{(x + 5)^2 }} \\ <br /> \end{array}<br /> \]<br />

 

[Mentallic]

log y = log [ (x-3)² / x]

I'm not sure about is how they got rid of the "log" to get only x and y in the answer, though. Are you even allowed to divide "log" and cross it out?

No you can't do that. Log is the inverse of exponentiating, in other words, if y=e^x then x=log_e(y)=ln(y). This means that when you take the exponential of a logarithm, you end up back where you started.
x=e^{ln(x)}

So if you take the exponential of both sides (not the same as dividing both sides by log, which doesn't make any sense) then the logs will cancel.

edit: since the question says to also state the domain, then I guess I should have mentioned this now crucial piece of information. It is not entirely true that x=e^{ln(x)} because logs are only defined for x>0, so you need to place the restriction down that x>0. If you have log(x-5) then x-5>0 or x>5 etc.

 

[symbolipoint]

Number b,

Equivalent to \[<br /> \log (y) + \log (1000) = \log (y + 1) + \log (x)<br /> \]<br />

\[<br /> \begin{array}{l}<br /> \log (y) - \log (y + 1) = \log (x) - \log 1000 \\ <br /> \log \frac{y}{{y + 1}} = \log \frac{x}{{1000}} \\ <br /> \frac{y}{{y + 1}} = \frac{x}{{1000}} \\ <br /> \end{array}<br /> \]<br />

Continuing with appropriate steps, you obtain the answer shown in your first post.

 

[HallsofIvy]

If you have f(x)= f(y) with "f" a function you cannot "divide by f"- you divide by numbers not functions. But if you know that a function is "one to one" (and logarithms are) then from "f(x)= f(y)" you can conclude "x= y".