Simplifying $3\log x + 0.5\log y - 2\log z$. I'm quite new to logs and exponentials and not particularly familiar with their 'rules' so you will probably have to explain quite deeply where I've gone wrong - hopefully this will also make it very useful to future readers of this question.
My attempt: 
$3\log x = \log x^3$ and so on so...
$\log x^3 + \log y^{0.5} - \log z^2$
and as $\log x^3$ and $\log y^3$ are to the same base...
$\log(x^3y^{0.5}) - \log z$
So my final attempt answer would be $\log(x^3y^{0.5})/\log z^2$.
Is this correct? Could it be further simplified? Or is it just wrong?
Thanks
 A: The last step is incorrect.  The rule is $$\log a - \log b = \log \frac{a}{b}$$
Note that there should be just one $\log$ function in the end.  Otherwise the solution is good.
A: Almost ok:
$$\log(x^3y^{0.5}) - \log z^2 = \log \left(\frac{x^3y^{0.5}}{z^2} \right).$$
A: It is just $$\log(x^3)+\log(y^{1/2})-\log(z^2)=\log\left(\frac{x^3y^{1/2}}{z^2}\right)$$
A: While your notation is quite hard to read, I think you mean to simplify
$$3\log(x)+\frac{1}{2}\log(y)-2\log(z),$$
where $\log$ is the natural logarithm. Or any other logarithm, it doesn't matter as the steps are the same. Then, assuming $x,y,z>0$, we can indeed put the coefficients inside the $\log$'s, turning them into exponents:
$$\log(x^3) + \log(\sqrt{y}) - \log(z^2).$$
Then, using the sum and difference formulas for logarithms (wikipedia), we get
$$\log\left(\frac{x^3\sqrt{y}}{z^2}\right)$$
So you where almost there. Note that you turned a minus into a 'divide by', which is illegal. The minus turns into a 'divide by' inside the $\log$
