Animated Solution for Mathematics - Binomial Theorem: If the fourth term in the expansion of $(x+x^{{\log }_{2}x}{)}^7$ is 4480, then the value of $x$ where $x\in \mathbb{N}$ is equal to :

Unmasking the Hidden Variable

My dear student, welcome to the arena. Today, we are not just solving a binomial expansion; we are embarking on a detective mission.

When you look at the expression $(x+x^{{\log }_{2}x}{)}^7$, what do you see? Do you see a terrifying monster of algebra, or do you see a beautiful, structured dance of powers? Let us choose the latter.

In the JEE Advanced, the difference between a student who struggles and a student who succeeds is the ability to remain calm when the expression looks unfamiliar. Let us peel back the layers of this problem together.

The General Term Toolkit

Every binomial expansion problem has a heartbeat, and that heartbeat is the general term formula:

In our case, $n=7$, $a=x$, and $b=x^{{\log }_{2}x}$. We are hunting for the 4th term, which means $r=3$.

When we plug these into our formula, we get:

Notice how the structure holds firm. We have the binomial coefficient ${}^7{C}_3$, which is simply:

This is our anchor. Now, we are left with $35\cdot x^4\cdot x^{3{\log }_{2}x}=4480$. The fog is starting to clear, isn't it?

The Logarithmic Bridge

Now, we arrive at the crux of the problem. We have $x^4\cdot x^{3{\log }_{2}x}=\frac{4480}{35}$.

A quick calculation tells us that $4480$ divided by $35$ is $128$. And here is where your intuition must kick in.

$128$ is not just a number; it is a power of $2$, specifically $2^7$. So, our equation becomes:

We have a variable in the base and a variable in the exponent. This is the classic 'logarithmic trap.'

How do we bring that exponent down to earth? We take the logarithm! By applying ${\log }_2$ to both sides, we transform the exponential mountain into a linear valley.

The Quadratic Revelation

Applying the power rule ${\log }_a(m^n)=n{\log }_{a}m$, our equation transforms into:

Suddenly, the complexity vanishes. If we let $t={\log }_{2}x$, we are staring at a simple quadratic equation:

This is the moment of triumph. We factorize this into $(3t+7)(t-1)=0$.

We find two potential values for $t$: $t=1$ and $t=-\frac{7}{3}$.

The Final Constraint

We are almost at the finish line. We have $t=1\Rightarrow {\log }_{2}x=1\Rightarrow x=2^1=2$.

And we have $t=-\frac{7}{3}\Rightarrow {\log }_{2}x=-\frac{7}{3}\Rightarrow x=2^{-\frac{7}{3}}$.

But wait! The problem explicitly states that $x\in \mathbb{N}$. This is the final gatekeeper of the JEE.

We must reject the non-natural solution. Thus, $x=2$ is our only champion.

You see? It wasn't about brute force; it was about structure, recognition, and respecting the constraints. You have mastered this. Keep this clarity, and you will conquer any problem the exam throws at you.