Animated Solution for Mathematics - Functions: If the domain of the function $f(x)={\log }_7(1-{\log }_4(x^2-9x+18))$ is $(\alpha ,\beta )\cup (\gamma ,\delta )$, then $\alpha +\beta +\gamma +\delta$ is equal to

Welcome, fellow explorers of the mathematical universe. Today, we are going to dissect a problem that often trips up even the most seasoned JEE aspirants: the domain of a nested logarithmic function.

When you see a function like $f(x)={\log }_7(1-{\log }_4(x^2-9x+18))$, it is easy to feel overwhelmed. Think of this function not as a single, terrifying entity, but as an onion. To find the domain, we simply need to peel it, layer by layer, from the inside out.

Let us start with the innermost part of our function: the quadratic expression $x^2-9x+18$. This expression is trapped inside a logarithm with base $4$. The fundamental rule of logarithms is that the argument must be strictly positive.

Therefore, our first constraint is:

To solve this, we factorize the quadratic. We are looking for two numbers that multiply to $18$ and add to $-9$. Those numbers are $-6$ and $-3$.

So, we have:

Using the wavy curve method, we identify the critical points at $x=3$ and $x=6$. Since we want the expression to be greater than zero, we look for the regions outside these roots. Thus, our first valid interval is $x\in (-\infty ,3)\cup (6,\infty )$.

Now, let us move to the outer layer. The entire argument of the ${\log }_7$ function must also be strictly positive. This gives us our second constraint:

This simplifies to:

Now, we apply the definition of a logarithm to remove the base $4$. Because $4>1$, the inequality sign remains unchanged. We get:

Subtracting $4$ from both sides, we arrive at a new quadratic inequality:

Factorizing this, we need two numbers that multiply to $14$ and add to $-9$. Those are $-2$ and $-7$. So, we have:

For this inequality, since we want the expression to be less than zero, the solution lies between the roots. Therefore, our second valid interval is $x\in (2,7)$.

We have two constraints. The first one, from the inner log, is $x\in (-\infty ,3)\cup (6,\infty )$. The second one, from the outer log, is $x\in (2,7)$.

The true domain of our function is the intersection of these two sets. We need to find where they overlap:

Let us visualize this on the number line. The first set covers everything less than $3$ and everything greater than $6$. The second set covers everything between $2$ and $7$.

They overlap between $2$ and $3$, and again between $6$ and $7$. Thus, the domain is $(2,3)\cup (6,7)$.

The problem states that the domain is $(\alpha ,\beta )\cup (\gamma ,\delta )$. By comparing our result $(2,3)\cup (6,7)$ with this format, we identify our variables: $\alpha =2$, $\beta =3$, $\gamma =6$, and $\delta =7$.

The question asks for the sum $\alpha +\beta +\gamma +\delta$. Calculating this:

The final answer is 18.