Find Domain of Log Function Need help determining the Domain of $$\log_2\left(2x+3\right)=\log_2\left(3x+4\right)-\log_2\left(3x+2\right)$$

Do not know if I should simplify first to get the domain, or to get the domain from the set logs that are in place?
This is an equation, and you can solve it. First of all we need to understand the single domains. That is easy. I will use the writing $\lg$ for the log base $2$.

$$\lg(2x+3)$$

This function exists iff $2x+3 &gt;0$ hence $\boxed{x &gt; -\frac{3}{2}}$.

$$\lg(3x+4)$$

This function exists if $3x+4 &gt;0$ hence $\boxed{x &gt; -\frac{4}{3}}$

$$\lg(3x+2)$$

This function exists if $3x+2 &gt;0$ hence $\boxed{x &gt;-\frac{2}{3}}$

Finally you can write th equation

$$\lg(2x+3) = \lg\left[(3x+4)/(3x+2)\right]$$

Where I used the well known property of the sum of log.

So now you can exponentiate in base $2$:

$$2x+3 = \frac{3x+4}{3x+2}$$

$$(2x+3)(3x+2) = 3x+4$$

$$6x^2 + 13x + 6 = 3x + 4$$

$$6x^2 + 10x + 2 = 0$$

Now you can solve it as a normal second degree equation

$$x = \frac{-10\pm\sqrt{100 - 48}}{12}$$ 

And check for the solutions, that don't have to be in contrast with the previous assumptions (necessarily).
I see an equation, not (just) a function?

In any case: $\log_a x$ (for a positive base $a$, different from $1$) is only defined for $x&gt;0$. So for any logarithm appearing in your equation, the argument of the logarithm has to be strictly positive.

E.g. for $\log_2(3x+2)$ you need $3x+2&gt;0 \Leftrightarrow x &gt; -\tfrac{2}{3}$.

For all logarithms to be defined, you'll want the intersection of the 'domains' (sets where the different logarithms are defined), before simplifying (see below).



Notice that
$$\log x + \log x$$
is defined only for $x&gt;0$ while (use $\log a + \log b = \log (ab)$)
$$\log \left( x^2 \right)$$
is defined for all $x\ne0$.