Basic understanding of Log and $2 \log _3(x)+\log _9(x)=10$ So this is what I have done so fare 
$$2 \log _3(x)+\log _9(x)=10$$
I know that 
$$\log _9(x)=\log _3\left(\sqrt{x}\right)$$
I therefore have
$$\log _3\left(x^{5/2}\right)=10$$
However here is where I realise that I have not properly understood the underlying of log.
Since I know that the answer is 81 I was able to realise that 
$$\log _3(x)=4$$
Is the same as x==3^4, but  I do not understand if Im dividing or multiplying or what is going on. I thought that 
$$\log _3(x)$$
meant that
$$3^?=x$$
Basically trying to find the exponent? At least that is the dummed version that I got thought in class
Could someone explain what is going on and what I do not understand , in such a way that I am able to grasp the concept.
 A: You correctly got:$$\log_3(x^{5/2})=10$$next step gives you:$$x^{5/2}=3^{10}$$therefore:$$x=(3^{10})^{2/5}$$Hopefully you can complete from here
A: $$\log _3\left(x^{5/2}\right)=10$$
I will use the approach that you stated to find the answer. 
You are on the right track when you state that $\log_3(x)$ is defined to be the solution to the equation $3^?=x$ where we solve for question-mark. In our case, the question mark is already known and we have to solve for $x$. For the equation above, question-mark equals 10. That's what the equals sign tells you in this scenario.
Now, we have to solve the following equation for $x$.
$$3^{10} = x^{5/2}$$
A: Basically, you have a function $f:\mathbb{R}\to\mathbb{R}^+$ defined by:
$$ f(x)=3^x.$$
Since $f$ is increasing and surjective, there exists a function $g:\mathbb{R^+}\to \mathbb{R}$ that is the inverse of $f$, i.e. satisfies:
$$\forall x\in\mathbb{R},\quad g(f(x))=x,$$
$$\forall x\in\mathbb{R}^+,\quad f(g(x))=x.$$
We call such function $\log_3$, i.e. $\log_3(x)=g(x)$. So yes, if $x\in\mathbb{R}^+$ and 
$$ 3^{?}=x, $$
we have:
$$ ? = \log_3(x).$$
All the algebraic properties of a logarithm function can be derived from the properties of its inverse function. For instance, $f(x)f(y)=f(x+y)$ gives $g(xy)=g(x)+g(y)$. This also gives that if you pick a "reference logarithm" as the inverse function of $e^x$ and you call it simply $\log x$, then:
$$\log_a(x) = \frac{\log x}{\log a},$$
so:
$$\log_{a^2}(x) = \frac{\log x}{\log a^2} = \frac{\log x}{2\log a} = \frac{1}{2}\log_a(x).$$
This properties give that your equation is equivalent to:
$$\left(2+\frac{1}{2}\right)\log_3(x) = 10, $$
or to:
$$\log_3(x) = 4,$$
so $x=3^4$ as you correctly stated.
A: Perhaps if you cleared out the denominator first it would be clearer what is going on:
$2\log_{3}x+\frac{1}{2}\log_3{x}=10\\
4\log_{3}x+\log_3{x}=20\\
5\log_{3}x=20\\
\log_3 x = 4\\
x=3^4$
Does this help?
