<s>[INST] Find $m$ such that the inequality has $4$ integer solutions. 
Find all values of $m$ such that the inequality: $\log(60x^2+120x+10m-10)>1+3\log(x+1)$ has exactly $4$ integer solutions?

The first thing I did was to gather from the inequality;
$\log(60x^2+120x+10m-10)>1+3\log(x+1)$
$\Leftrightarrow 1+\log(6x^2+12x+m-1)>1+\log(x^3+3x^2+3x+1)$
$\Leftrightarrow 6x^2+12x+m-1>x^3+3x^2+3x+1$
$\Leftrightarrow 0>x^3-3x^2-9x-m+2$
Right here, I have no idea of what to do next, any help would be appreciated. Thank you!
 [/INST] Consider the inequality
$$m - 2 > x^3 - 3x^2 -  9x. \tag1 $$
As $~x \to -\infty, ~$ the RHS of (1) above goes to $-\infty.$
Therefore, it is tempting, but wrong to assume that regardless of the (fixed) value of $m$ chosen, there will be an infinite number of (negative) integer solutions.
The reason that this is wrong is that the original problem involves logarithms, and you can not take the logarithm of a non-positive number.  Therefore, there are two additional (hidden) constraints:

*

*$(x + 1) > 0 \iff x > -1$

*$(60x^2+120x+10m-10) > 0 $ 
$\iff (6x^2 + 12x + (m-1) > 0$ 
$\iff (36x^2 + 72x + (6m - 6) > 0 $ 
$\iff (6x + 6)^2 + (-36 + 6m - 6) > 0 $ 
$\iff [6(x + 1)]^2 > 42 - 6m \iff $ 
$\iff 6 \times [(x + 1)^2] > 7 - m$ 
$\iff m > 7 - \{6 \times [(x + 1)^2]\}$.

Further, given the problem's preCalculus tag, only purely algebraic methods of attack are permitted.
Let $f(x) = x^3 - 3x^2 - 9x.$  Then, the range of $m$ must be identified so that there will be exactly $4$ integer values that satisfy all of the following constraints:

*

*$(m - 2) > f(x).~$ : Constraint-1

*$x > -1.~$ : Constraint-2

*$m > 7 - \{6 \times [(x + 1)^2]\}.~$ : Constraint-3

Then, considering Constraint-2 above, 
$f(0) = 0.$ 
$f(1) = -11$. 
$f(2) = -22$. 
$f(3) = -27$. 
$f(4) = -20$. 
$f(5) = 5$. 
$f(6) = 54.$
Edit
For $x \geq 6,$ you have that 
$f(x+1) - f(x) = (3x^2 + 3x + 1) - 3(2x + 1) - 9$ 
$= ~3x^2 - 3x - 11 = 3(x^2 - x) - 11 > 0.$ 
Therefore, for $x \in \Bbb{Z_{\geq 6}}, f(x)$ is strictly increasing.
A natural try is to explore what happens if 
$(m - 2) > 0$.  
You would also need 
$m > 7 - \{6 \times [(x + 1)^2]\}$.
For $x \geq 0$, this will automatically be satisfied if $m > 1.$
Therefore, if $(m - 2) = 0$ then you will have the $4$ solutions of $x \in \{1,2,3,4\}.$
So, the question is, what happens if $m < 2$, or $m > 2 ~$ ?  $~~m > 2$ must be excluded, since that would automatically add the solution $x = 0$.
Note that $f(1) = -11.$ 
If $(x = 1)$, then you need that 

*

*$(m - 2) > -11$ 

*$m > 7 - \{6 \times [(2)^2]\} = -17.$
Therefore, the preliminary answer seems to be
$$-9 < m \leq 2 \tag2 $$.
Note that if $m \leq -9$, then $x = 1$ is not a solution, $x = 0$ is not a solution, and $x \geq 5$ is not a solution.  Therefore, the range given in (2) above is the final answer.
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