Key Concept: Binomial Theorem - General Term

The binomial theorem provides a formula for the expansion of powers of a binomial $(a+b)^n$. The general term, often denoted as $T_{r+1}$, gives any specific term in this expansion.

The formula for the $(r+1)^{th}$ term in the binomial expansion of $(a+b)^n$ is: $T_{r+1} = {}^nC_r \cdot a^{n-r} \cdot b^r$ where ${}^nC_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

In this problem, we are given the binomial expression ${\left( {{2 \over x} + {x^{{{\log }_8}x}}} \right)^6}$. Comparing this with $(a+b)^n$, we have:

• $a = {2 \over x}$
• $b = {x^{{{\log }_8}x}}$
• $n = 6$

We are asked to find a value of $x$ such that the fourth term ($T_4$) of this expansion is equal to $20 \times 8^7$.

Calculating the Fourth Term ($T_4$)

To find the fourth term, we set $r+1 = 4$, which implies $r = 3$. Now, substitute $n=6$, $r=3$, $a = {2 \over x}$ and $b = {x^{{{\log }_8}x}}$ into the general term formula: $T_4 = T_{3+1} = {}^6C_3 \cdot {\left( {{2 \over x}} \right)^{6-3}} \cdot {\left( {{x^{{{\log }_8}x}}} \right)^3}$

Let's calculate each part:

1. Binomial Coefficient: ${}^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

2. First Term Power: ${\left( {{2 \over x}} \right)^{6-3}} = {\left( {{2 \over x}} \right)^3} = \frac{2^3}{x^3} = \frac{8}{x^3}$

3. Second Term Power: Using the exponent rule $(m^p)^q = m^{pq}$: ${\left( {{x^{{{\log }_8}x}}} \right)^3} = x^{3 \cdot {{\log }_8}x}$

Now, combine these parts to get the expression for $T_4$: $T_4 = 20 \cdot \frac{8}{x^3} \cdot x^{3{{\log }_8}x}$

Setting Up the Equation

We are given that $T_4 = 20 \times 8^7$. So, we equate our derived expression for $T_4$ to the given value: $20 \cdot \frac{8}{x^3} \cdot x^{3{{\log }_8}x} = 20 \times 8^7$

To simplify, we can divide both sides by 20: $\frac{8}{x^3} \cdot x^{3{{\log }_8}x} = 8^7$

Now, we rewrite $\frac{8}{x^3}$ as $8 \cdot x^{-3}$ and use the exponent rule $m^p \cdot m^q = m^{p+q}$: $8 \cdot x^{-3} \cdot x^{3{{\log }_8}x} = 8^7$ $8 \cdot x^{(3{{\log }_8}x - 3)} = 8^7$

Divide both sides by 8: $x^{(3{{\log }_8}x - 3)} = \frac{8^7}{8^1}$ $x^{(3{{\log }_8}x - 3)} = 8^{7-1}$ $x^{(3{{\log }_8}x - 3)} = 8^6$

Solving for x using Logarithms

To solve for $x$ when it appears in the exponent and also as the base of a logarithm, we take the logarithm of both sides. Since the base of the logarithm in the exponent is 8, it is strategic to take $\log_8$ on both sides: $\log_8 \left( x^{(3{{\log }_8}x - 3)} \right) = \log_8 (8^6)$

Apply the logarithm property $\log_b (M^P) = P \log_b M$ to the left side and $\log_b b^P = P$ to the right side: $(3{{\log }_8}x - 3) \cdot {{\log }_8}x = 6$

This equation looks complicated, but we can simplify it by making a substitution. Let $y = {{\log }_8}x$. Substituting $y$ into the equation: $(3y - 3)y = 6$

Distribute $y$ on the left side: $3y^2 - 3y = 6$

Rearrange the equation into a standard quadratic form $(ay^2 + by + c = 0)$ by moving all terms to one side: $3y^2 - 3y - 6 = 0$

Divide the entire equation by 3 to simplify: $y^2 - y - 2 = 0$

Now, factor the quadratic equation. We need two numbers that multiply to -2 and add up to -1. These numbers are -2 and 1: $(y - 2)(y + 1) = 0$

This gives us two possible values for $y$: $y - 2 = 0 \Rightarrow y = 2$ $y + 1 = 0 \Rightarrow y = -1$

Finally, substitute back $y = {{\log }_8}x$ to find the values of $x$: Case 1: $y = 2$ ${{\log }_8}x = 2$ Using the definition of logarithm $( \log_b M = P \Rightarrow M = b^P )$: $x = 8^2$ $x = 64$

Case 2: $y = -1$ ${{\log }_8}x = -1$ Using the definition of logarithm: $x = 8^{-1}$ $x = \frac{1}{8}$

Verification and Conclusion

The problem states that $x > 0$. Both our solutions, $x = 64$ and $x = \frac{1}{8}$, satisfy this condition. Therefore, the possible values of $x$ are $8^2$ and $8^{-1}$.

Looking at the provided options: (A) $8^{-2}$ (B) $8^2$ (C) $8^3$ (D) $8$

Among the options, $8^2$ is a valid value for $x$ (Option B). It's important to note a discrepancy: the problem states the correct answer is (A) $8^{-2}$. However, based on the calculation, $8^{-2}$ is not a solution. If the question or options were slightly different, $8^{-2}$ might be a solution, but with the given problem statement, $x = 8^2$ and $x = 8^{-1}$ are the mathematically derived solutions. If only one option must be chosen, $8^2$ is a correct possibility from our calculation.

Important Tips and Common Pitfalls

• Binomial Coefficient: Always calculate ${}^nC_r$ carefully.
• Exponent Rules: Be meticulous when combining terms with exponents, especially when dealing with negative exponents or powers of powers. Remember $(a^m)^n = a^{mn}$ and $a^m \cdot a^n = a^{m+n}$.
• Logarithm Properties: The property $\log_b (M^P) = P \log_b M$ is crucial for solving equations where the variable is in the exponent. Also, recall the definition of logarithm: $\log_b M = P \iff M = b^P$.
• Quadratic Equations: Ensure you correctly solve the resulting quadratic equation, either by factoring, using the quadratic formula, or completing the square.
• Domain Restrictions: Always check if your solutions satisfy any given domain restrictions (e.g., $x>0$ for logarithmic expressions).
• Double-Check: It's good practice to plug your solutions back into the original equation to ensure they hold true.

Key Takeaway

This problem effectively combines concepts from binomial theorem, exponent rules, and logarithms, leading to a quadratic equation. The key is to systematically apply the relevant formulas and properties, especially for handling exponential and logarithmic terms, to simplify the equation to a solvable form. Always be careful with arithmetic and algebraic manipulations.