Solve $\log x + \log(2x-5) = \log 96 – \log 8$ 
Solve $\log x + \log (2x-5) = \log 96 – \log 8$.

I started by doing this:
$\log 2x^2-5x = \log 12$
Then I got rid of the logs. Can I do that?
$2x^2-5x = 12$
What do I do next? What is $x$?
 A: If $\log x+\log(2x-5)=\log 96–\log 8$, then $\log (2x^2-5x)=\log 12$ so that $2x^2-5x=12$. Write this as $(2x+3)(x-4)=2x^2-5x-12=0$ so that $x=4$ or $x=-\frac{3}{2}$. However the latter is not a solution as one requires $x>0$ since we speak of $\log x$.
A: Put your equation in the form $$ax^2+bx+c=0$$ and apply the general quadratic equation formula$$OR$$
Try to factorize the equation into the form $$(a_1x+b_1)(a_2x+b_2)=0$$ and solve.
NB Note that $\log x \in \mathbb R \iff x>0$ so you only take positive values of $x$.
A: Use some fundamental theorems/lemmas$^{1,2}$ for the simplification.
$\log(x) +\log(2x + 5) = \log(96)-\log(8)$ which simplifies to $\log(2x^2 + 5x) = \log(12)$. You can get rid of the $\log$ operator as both sides are equal. We have $2x^2 + 5x = 12 \, \, \, \, \Leftrightarrow \, \, \, \, 2x^2 + 5x - 12 = 0 $. Solve the quadratic in $x$ and eliminate the negative solutions since $\log(x)$ is defined when $x$ is strictly positive.

$^1$ $\log(a\cdot b) = \log(a)+\log(b)$. $^2$$\log(a\div b)=\log(a) - \log(b)$
