If the fourth term in the binomial expansion of (2/x + x^(log₈ x))⁶ (x 0) is 20 × 8⁷, then a value

Q: Wait, why do we reject the negative root or do we keep both? The quadratic gave me $t=6$ and $t=-3$.

That is a fantastic observation! In a purely mathematical context, both $t=6$ and $t=-3$ are valid solutions for the quadratic equation. This leads to $x = 2^6 = 64$ and $x = 2^{-3} = 1/8$. However, in a competitive exam like JEE, we must look at the provided options. Since $64$ is $8^2$ and $1/8$ is $8^{-1}$, we check which one is listed. Only $8^2$ is an option, so we select that. Always let the options guide your final selection!

Q: Why can't we just use energy conservation or some other shortcut? This binomial expansion feels like a long road.

I hear your frustration! It feels like a long road, but in algebra, there are no shortcuts for the fundamental structure. The binomial expansion is the 'map' here. By following the general term formula, we systematically break down the complexity. The 'shortcut' is actually the algebraic simplification—once you see the quadratic structure, the problem collapses quickly. Trust the process!

Q: Is the change of base property really necessary? Can't I just keep it as $\log_8 x$?

You technically could, but it would make the algebra significantly harder. By converting to base 2, we utilize the fact that $8 = 2^3$. This allows the $3$ in the exponent to cancel out with the denominator from the change of base formula. It is a strategic move to make the numbers 'talk' to each other. Always look for ways to make the bases match!

Question

If the fourth term in the binomial expansion of $\left( \frac{2}{x} + x^{\log_8 x} ight)^6 (x > 0)$ is $20 	imes 8^7$, then a value of $x$ is :

Options:

(A) 8

(B) $8^2$

(C) $8^{-2}$

(D) $8^3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every binomial expansion problem has a North Star: the General Term formula. For any expression $(a + b)^n$, the $(r+1)$-th term is defined as:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

In our case, we are looking for the fourth term, which means $r+1 = 4$, so $r = 3$. Our parameters are $n = 6$, $a = \frac{2}{x}$, and $b = x^{\log_8 x}$.

By substituting these into our formula, we get:

$$T_4 = \binom{6}{3} \left( \frac{2}{x} ight)^{6-3} \left( x^{\log_8 x} ight)^3$$

The binomial coefficient $\binom{6}{3}$ is a constant, calculated as:

$$\binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$

### The Algebraic Dance

Now, let us simplify the expression. We have $T_4 = 20 \cdot (\frac{2}{x})^3 \cdot (x^{\log_8 x})^3$.

Expanding the first part, $(\frac{2}{x})^3$ becomes $\frac{8}{x^3}$, or $8x^{-3}$. The second part, $(x^{\log_8 x})^3$, becomes $x^{3 \log_8 x}$.

Multiplying these, we get:

$$T_4 = 160 \cdot x^{3 \log_8 x - 3}$$

We are told this term equals $20 	imes 8^7$. Setting them equal, we have $160 \cdot x^{3 \log_8 x - 3} = 20 	imes 8^7$.

Since $20 	imes 8 = 160$, we can divide both sides by $160$, leaving us with:

$$x^{3 \log_8 x - 3} = 8^6$$

### The Logarithmic Bridge

The base $8$ on the right and the $\log_8$ in the exponent suggest a change of base. Let us use the property $\log_8 x = \frac{\log_2 x}{\log_2 8}$.

Since $8 = 2^3$, we know $\log_2 8 = 3$. Thus, $\log_8 x = \frac{\log_2 x}{3}$.

Substituting this back into our exponent, we get $3 \cdot (\frac{\log_2 x}{3}) - 3$, which simplifies to $\log_2 x - 3$. Our equation is now:

$$x^{\log_2 x - 3} = 8^6$$

### The Quadratic Climax

To solve for $x$, we take $\log_2$ on both sides. Note that $8^6 = (2^3)^6 = 2^{18}$.

Taking the log:

$$\log_2 \left( x^{\log_2 x - 3} ight) = \log_2 \left( 2^{18} ight)$$

Using the power rule for logarithms, the exponent comes down:

$$(\log_2 x - 3) \cdot \log_2 x = 18$$

Let $t = \log_2 x$. We have arrived at the classic quadratic form:

$$t^2 - 3t - 18 = 0$$

Factoring this, we get $(t-6)(t+3) = 0$. This gives us two potential paths: $t = 6$ or $t = -3$.

If $t = 6$, then $\log_2 x = 6$, which means $x = 2^6 = 64$. If $t = -3$, then $\log_2 x = -3$, which means $x = 2^{-3} = \frac{1}{8}$.

The final solutions are **$x = 64$** or **$x = \frac{1}{8}$**.

Animated Solution for Mathematics - Binomial Theorem: If the fourth term in the binomial expansion of $(\frac{2}{x} + x^{l o g_{8} x})^{6} (x > 0)$ is $20 \times 8^{7}$ , then a value of $x$ is :

Select Answer:

Visualized Solution

General Term of Binomial Expansion

Substituting Values into $T_{4}$

Simplifying the Binomial Coefficient

Expanding the Terms

Combining Exponents

Equating to the Given 4th Term

Canceling the Constants

Applying Change of Base Property

Substituting the Logarithm

Taking Logarithm on Both Sides

Forming a Quadratic Equation

Factoring the Quadratic

Finding the Values of $x$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Algebraic Dance

The Logarithmic Bridge

The Quadratic Climax

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