Question

The complete solution set of the inequality $\frac{1}{\log_4(\frac{x+1}{x+2})} > \frac{1}{\log_4(x+3)}$ is $(\frac{-a}{20}, \infty)$, then determine $a$

Answer 1

20

Answer 2

The given inequality is $\frac{1}{\log_4(\frac{x+1}{x+2})} > \frac{1}{\log_4(x+3)}$.

First, we determine the domain of the inequality.

1. The arguments of the logarithms must be positive:
   $\frac{x+1}{x+2} > 0$ and $x+3 > 0$.
   $\frac{x+1}{x+2} > 0$ holds for $x \in (-\infty, -2) \cup (-1, \infty)$.
   $x+3 > 0$ holds for $x \in (-3, \infty)$.
   The intersection of these two conditions is $x \in (-3, -2) \cup (-1, \infty)$.

2. The denominators cannot be zero:
   $\log_4(\frac{x+1}{x+2}) \neq 0 \implies \frac{x+1}{x+2} \neq 4^0 = 1 \implies x+1 \neq x+2$, which is always true.
   $\log_4(x+3) \neq 0 \implies x+3 \neq 4^0 = 1 \implies x \neq -2$. This is already excluded by the domain condition $\frac{x+1}{x+2} > 0$.

So, the domain of the inequality is $x \in (-3, -2) \cup (-1, \infty)$.

Let $A = \log_4(\frac{x+1}{x+2})$ and $B = \log_4(x+3)$. The inequality is $\frac{1}{A} > \frac{1}{B}$.
This is equivalent to $\frac{1}{A} - \frac{1}{B} > 0 \implies \frac{B-A}{AB} > 0$.

The inequality $\frac{B-A}{AB} > 0$ holds if ($B-A > 0$ and $AB > 0$) or ($B-A < 0$ and $AB < 0$).

Sign of $A = \log_4(\frac{x+1}{x+2})$:
$A > 0 \iff \frac{x+1}{x+2} > 1 \iff \frac{-1}{x+2} > 0 \iff x+2 < 0 \iff x < -2$.
$A < 0 \iff 0 < \frac{x+1}{x+2} < 1 \iff \frac{-1}{x+2} < 0 \iff x+2 > 0 \iff x > -2$.

Sign of $B = \log_4(x+3)$:
$B > 0 \iff x+3 > 1 \iff x > -2$.
$B < 0 \iff 0 < x+3 < 1 \iff -3 < x < -2$.

Sign of $B-A = \log_4(x+3) - \log_4(\frac{x+1}{x+2}) = \log_4\left(\frac{(x+3)(x+2)}{x+1}\right)$.
$B-A > 0 \iff \frac{(x+3)(x+2)}{x+1} > 1 \iff \frac{x^2+5x+6 - (x+1)}{x+1} > 0 \iff \frac{x^2+4x+5}{x+1} > 0$.
The quadratic $x^2+4x+5$ has discriminant $4^2 - 4(1)(5) = -4 < 0$ and leading coefficient 1 > 0, so $x^2+4x+5 > 0$ for all real $x$.
Thus, $B-A > 0 \iff x+1 > 0 \iff x > -1$.
$B-A < 0 \iff x+1 < 0 \iff x < -1$.

Now let's check the conditions for $\frac{B-A}{AB} > 0$ in the two parts of the domain:

Case 1: $x \in (-3, -2)$.
In this interval:
$x < -2$, so $A > 0$.
$-3 < x < -2$, so $B < 0$.
$AB < 0$.
$x < -1$, so $B-A < 0$.
The condition $\frac{B-A}{AB} > 0$ becomes $\frac{\text{negative}}{\text{negative}} > 0$, which is true.
So, the interval $(-3, -2)$ is part of the solution set.

Case 2: $x \in (-1, \infty)$.
In this interval:
$x > -1$, so $x > -2$, which means $A < 0$.
$x > -1$, so $x > -2$, which means $B > 0$.
$AB < 0$.
$x > -1$, so $B-A > 0$.
The condition $\frac{B-A}{AB} > 0$ becomes $\frac{\text{positive}}{\text{negative}} > 0$, which is false.
So, the interval $(-1, \infty)$ is not part of the solution set.

The complete solution set is $(-3, -2)$.

The problem states that the complete solution set is $(\frac{-a}{20}, \infty)$. This form suggests that the solution set is a single interval extending to infinity. This contradicts the derived solution set $(-3, -2)$.

If the intended inequality was $\frac{1}{\log_4(\frac{x+1}{x+2})} < \frac{1}{\log_4(x+3)}$ instead of $>$, the solution set would be $(-1, \infty)$. In this case, we have $-1 = \frac{-a}{20}$, which implies $a = 20$.

Therefore, assuming the intended inequality was $\frac{1}{\log_4(\frac{x+1}{x+2})} < \frac{1}{\log_4(x+3)}$, the solution set is $(-1, \infty)$. Comparing with $(\frac{-a}{20}, \infty)$, we get $\frac{-a}{20} = -1$, which gives $a = 20$.