If the fourth term in the binomial expansion of (√(1/(10^(1+log10x))) + 12√(x^6))^6 is equal to 200,

Q: Wait, why do we reject the negative root $t = -4$?

Great catch! Remember that we defined $t = \log_{10}x$. The problem explicitly states that $x > 1$. If $x > 1$, then $\log_{10}x$ must be greater than $\log_{10}1$, which is $0$. Therefore, $t$ must be positive. Since $-4$ is negative, it violates the domain constraint of the problem.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure mathematics problem, specifically from the domain of Algebra. While physics problems often use binomial approximations, this is an exact binomial expansion. We rely on the Binomial Theorem, not physical laws, to find the term.

Q: Is the expression in the question actually correct?

You are very observant! The expression provided in the prompt had a slight typo. In competitive exams like JEE, you must be prepared to identify and correct such errors. The corrected form $(x^{\frac{1}{2(1+\log_{10}x)}} + x^{\frac{1}{12}})^6$ is the standard, solvable version of this problem.

Question

If the fourth term in the binomial expansion of $(\sqrt{1/(10^{1+\log_{10}x})} + \sqrt[12]{x^6})^6$ is equal to 200, and $x > 1$, then the value of x is :

Options:

(A) $10^3$

(B) $100$

(C) $10^4$

(D) $10$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The beauty of the Binomial Theorem lies in its ability to decompose complex expressions into manageable components. We are tasked with finding the value of $x$ given that the fourth term of the expansion of $\left( x^{\frac{1}{2(1+\log_{10}x)}} + x^{\frac{1}{12}} \right)^6$ is equal to $200$.

For any binomial expression $(a+b)^n$, the general term is defined as:

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

We are interested in the fourth term, $T_4$. By setting $r+1 = 4$, we identify $r = 3$ and $n = 6$.

### The Master Equation

Substituting our values into the general term formula, we get:

$$T_4 = \binom{6}{3} \left( x^{\frac{1}{2(1+\log_{10}x)}} \right)^{6-3} \cdot \left( x^{\frac{1}{12}} \right)^3$$

Calculating the binomial coefficient, we find $\binom{6}{3} = \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} = 20$. Setting the term equal to $200$, we have:

$$20 \cdot \left( x^{\frac{1}{2(1+\log_{10}x)}} \right)^3 \cdot \left( x^{\frac{1}{12}} \right)^3 = 200$$

Dividing both sides by $20$ simplifies the expression to:

$$x^{\frac{3}{2(1+\log_{10}x)}} \cdot x^{\frac{3}{12}} = 10$$

### Taming the Logarithmic Exponent

Simplifying the exponent $\frac{3}{12}$ to $\frac{1}{4}$, the equation becomes:

$$x^{\frac{3}{2(1+\log_{10}x)} + \frac{1}{4}} = 10$$

To solve for $x$, we let $t = \log_{10}x$ and take the logarithm base $10$ on both sides:

$$\left( \frac{3}{2(1+t)} + \frac{1}{4} \right) \cdot t = \log_{10}10 = 1$$

### The Algebraic Resolution

We now solve the resulting algebraic equation:

$$\frac{3t}{2(1+t)} + \frac{t}{4} = 1$$

Multiplying the entire equation by $4(1+t)$ to clear the denominators, we obtain:

$$6t + t(1+t) = 4(1+t)$$

Expanding and rearranging the terms leads to a quadratic equation:

$$t^2 + 3t - 4 = 0$$

Factoring the quadratic gives $(t+4)(t-1) = 0$, yielding roots $t = -4$ and $t = 1$. Given the constraint $x > 1$, we must have $t = \log_{10}x > 0$.

Therefore, we reject $t = -4$ and accept $t = 1$. Solving for $x$:

$$\log_{10}x = 1 \Rightarrow x = 10^1$$

The final value is **$x = 10$**.

Animated Solution for Mathematics - Binomial Theorem: If the fourth term in the binomial expansion of $(1/ (1 0^{1 + l o g_{10} x}) + 12 x^{6})^{6}$ is equal to 200, and $x > 1$ , then the value of x is :

Select Answer:

Visualized Solution

The Corrected Binomial Expression

General Term Formula

Identify Parameters for $T_{4}$

Extract Term $a$

Extract Term $b$

Substitute into $T_{4}$

Simplify the Equation

Combine Exponents

Take Logarithm on Both Sides

Form the Algebraic Equation

Clear Denominators

Form the Quadratic Equation

Factorize the Quadratic

Apply Constraints and Solve

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Master Equation

Taming the Logarithmic Exponent

The Algebraic Resolution

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