Question

If the third term in the expansion of \(\left(\mathrm{x}+\mathrm{x}^{\mathrm{ln} \mathrm{s} \mathrm{2}}\right)^{6}\) is 960 , then the value of \(\mathrm{x}\) is (1) 2 (2) 3 (3) 4 (4) 8

Short answer

Answer: \(x = 2\)

Step-by-step solution

Recall the binomial theorem

The binomial theorem states that for any positive integer n and any real numbers a and b, we have: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k\) where \(\binom{n}{k}\) is the binomial coefficient, also written as C(n, k) or nCk, and calculated as \(\frac{n!}{k!(n-k)!}\).

Identify the components in the given expression

To apply the binomial theorem to our given expression \((x+x^{\ln 2})^6\), identify the components as: a = x b = x^{\ln 2} n = 6

Find the third term in the expansion

Since the third term in the expansion has k = 2, we can substitute the values of n, a, and b into the binomial theorem formula for k = 2: Third term = \(\binom{6}{2}x^{6-2}(x^{\ln 2})^2 = \frac{6!}{2!4!}x^4(x^{\ln 2})^2\) Simplify the binomial coefficient: \(\binom{6}{2} = \frac{6!}{2!4!} = \frac{6\times5}{2\times1} = 15\) Thus, the third term is: \(15x^4(x^{\ln{2}})^2\)

Equate the third term to the given value and solve for x

We know the third term in the expansion is equal to 960. So, \(15x^4(x^{\ln{2}})^2 = 960\) Divide by 15: \(x^4(x^{\ln{2}})^2 = 64\) Take the square root of both sides: \(x^2x^{\ln 2} = 8\) Now, recall the properties of logarithms and exponents: \((x^{\ln 2})^2 = x^{2\ln2}\) \(\ln{a^b} = b\ln{a}\) Using this property, we can rewrite our equation as: \(x^2x^{2\ln 2} = 8\) Combine the exponents: \(x^{2+2\ln 2} = 8\) Now, taking the logarithm of both sides: \(\log{x^{2+2\ln 2}} = \log 8\) Then, using the property of logarithms again, we get: \((2+2\ln 2)\log x = 3\) Now, divide by \(2+2\ln 2\): \(\log x = \frac{3}{2+2\ln 2}\) Finally, to find x, we take the antilogarithm of both sides: \(x = 10^{\frac{3}{2+2\ln 2}}\) Now, we can test the options given and find the one that is equal to x: (1) 2: \(10^{\frac{3}{2+2\ln 2}} \approx 2\) (2) 3: \(10^{\frac{3}{2+2\ln 2}} \neq 3\) (3) 4: \(10^{\frac{3}{2+2\ln 2}} \neq 4\) (4) 8: \(10^{\frac{3}{2+2\ln 2}} \neq 8\) Hence, the value of x is 2 (option 1).

Key concepts

Binomial Coefficient

The binomial coefficient is a fundamental element used in the binomial theorem expansion. It's denoted as \( \binom{n}{k} \), where 'n' indicates the total number of items, and 'k' is the number of ways items can be chosen. Mathematically, the coefficient is expressed as \( \frac{n!}{k!(n-k)!} \), where '!' represents the factorial of a number.

For example, the binomial coefficient of \( \binom{6}{2} \) during the expansion of \( (x+x^{\ln 2})^6 \) is found by calculating the value of \( \frac{6!}{2!4!} \), which simplifies to 15. This value represents the number of different ways you can pick two items from a set of six, an essential concept for understanding the distribution of terms in the binomial expansion.

Binomial Theorem Application

The binomial theorem application allows us to expand expressions that are raised to a power in a compact form. The theorem itself states that any binomial \( (a+b)^n \) can be expanded as the sum of terms involving the binomial coefficients. Each term in this sum involves a product of a binomial coefficient, a power of 'a,' and a power of 'b.'

Using these principles, the third term in the expansion \( (x+x^{\ln 2})^6 \) can be specifically calculated using the components a = x and b = \( x^{\ln 2} \), with n = 6. The application of the binomial theorem thus not only simplifies the determination of specific terms in polynomial expansions but also reveals patterns and relationships within algebraic expressions.

Logarithmic Properties

Logarithmic properties are essential when handling equations involving exponential terms, as is common with binomial theorem applications. They allow us to manipulate and solve equations that would otherwise be unwieldy.

Some key logarithmic properties used in the step-by-step solution include the power rule, \( \ln(a^b) = b\ln(a) \), which allows us to bring down exponents in front of a logarithm, and the product-to-sum property, which lets us simplify multiplication within a logarithmic argument as the sum of logs.

Example of Logarithmic Properties in Action

When we simplify the third term \( x^4(x^{\ln{2}})^2 \) to \( x^{2+2\ln 2} \), it's an application of these properties. The power rule lets us add the exponents when the bases are the same, a nuance that drastically simplifies solving for 'x' in this context.