1. ## Laws of logs

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 (x^2/y) = 2 log(x + 5)

a) y = (x-3)^2 / x
b) y = x / (1000 -x)
c) y = x^2 / (x + 5)^2

I got the first one by isolating log y on the left side, then using the quotient law on the right side. I ended up with what they got for the answer, in addition to "log" in front of both sides, though. I'm not sure about is how they got rid of the "log" to get only x and y in the answer. Are you even allowed to divide "log" and cross it out?
I don't know where they got 1000 from in b).

2. Some books say Log is the Ln and the other it is base 10. I am not sure what your book states so I will discuss both which are identical in manner.

$\displaystyle \displaystyle log(y)=log\left(\frac{(x-3)^2}{x}\right)$

This is what we obtain through simplification. Here is when we should observe the domain as well. If x = 3, then we would have the log(0) = DNE so 3 is out and if x = 0, we would be dividing by 0 so 0 is out.

Domain:

$\displaystyle \displaystyle (-\infty, 0) \cup (0,3) \cup (3, \infty)$

Now to isolate the terms inside the log, we raise log base 10 to 10th power and ln to the e.

$\displaystyle \displaystyle 10^{log(y)}=10^{log\left(\frac{(x-3)^2}{x}}\right)}\rightarrow y=\frac{(x-3)^2}{x}$

Now you try the rest.

3. Thanks. We haven't learned raising logs to the nth power yet, or ln, but I sort of understand it now.

4. dwsnith misspoke when he referred to raising "log base 10 to the 10th power and ln to the e power". He meant taking 10 to the $\displaystyle log_{10}(x)$ power and e to the ln(x) power. And that comes from the definition of logarithm.

Fundamentally, it is just a matter of using the fact the logarithm function is one-to-one: if log(x)= log(y) then x= y.