Solution: Finding 'x' using the Binomial Theorem and Logarithmic Properties

This problem combines concepts from the Binomial Theorem and the properties of logarithms. We are asked to find a possible value of $x$ for which the ninth term in a given binomial expansion equals 180.

---

1. Key Concepts and Formulas

• Binomial Theorem: The general term (or $(r+1)$-th term) in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^n C_r a^{n-r} b^r$ where ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.
• Logarithm Property 1: $p^{\log_p M} = M$ This property allows us to simplify expressions where a base is raised to a logarithm with the same base.
• Logarithm Property 2: $k \log_p M = \log_p M^k$ This property allows us to move a coefficient into the exponent of the logarithm's argument.

---

2. Step-by-Step Working

Step 2.1: Simplify the terms within the binomial expression.

The given binomial expression is of the form $(A+B)^{10}$, where: $A = {3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }}$ $B = {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}$

Let's simplify $A$: $A = {3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }}$ Applying Logarithm Property 1 ($p^{\log_p M} = M$), with $p=3$ and $M = \sqrt {{{25}^{x - 1}} + 7}$: $A = \sqrt {{{25}^{x - 1}} + 7}$ This can also be written as $A = \left( {{25}^{x - 1}} + 7 \right)^{1/2}$.

Now, let's simplify $B$: $B = {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}$ First, apply Logarithm Property 2 ($k \log_p M = \log_p M^k$) to the exponent, with $k = -1/8$, $p=3$, and $M = ({5^{x - 1}} + 1)$: $B = {3^{{{\log }_3}({5^{x - 1}} + 1)^{ - {1 \over 8}} }}$ Now, apply Logarithm Property 1 ($p^{\log_p M} = M$), with $p=3$ and $M = ({5^{x - 1}} + 1)^{ - {1 \over 8}}$: $B = ({5^{x - 1}} + 1)^{ - {1 \over 8}}$

So, the binomial expression simplifies to $\left( \sqrt {{{25}^{x - 1}} + 7} + ({5^{x - 1}} + 1)^{ - {1 \over 8}} \right)^{10}$.

Step 2.2: Determine the ninth term of the expansion.

The problem states that the expansion is in "increasing powers of ${3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}$", which is our term $B$. This means $B$ is considered the second term in the binomial expansion $(A+B)^{10}$. To find the ninth term, $T_9$, we use the general term formula $T_{r+1} = {}^n C_r a^{n-r} b^r$. For the ninth term, $r+1 = 9$, which implies $r=8$. Here, $n=10$, $a=A$, and $b=B$. So, the ninth term $T_9$ is: $T_9 = {}^{10}C_8 A^{10-8} B^8 = {}^{10}C_8 A^2 B^8$

Step 2.3: Substitute the simplified terms into the expression for $T_9$ and set up the equation.

We need to calculate $A^2$ and $B^8$: $A^2 = \left( \sqrt {{{25}^{x - 1}} + 7} \right)^2 = {{25}^{x - 1}} + 7$ $B^8 = \left( ({5^{x - 1}} + 1)^{ - {1 \over 8}} \right)^8 = ({5^{x - 1}} + 1)^{ - 1} = \frac{1}{{{5^{x - 1}} + 1}}$

Next, calculate the binomial coefficient ${}^{10}C_8$: ${}^{10}C_8 = {}^{10}C_{10-8} = {}^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$

Now, substitute these into the expression for $T_9$: $T_9 = 45 \times \left( {{25}^{x - 1}} + 7 \right) \times \left( \frac{1}{{{5^{x - 1}} + 1}} \right)$ The problem states that the ninth term is equal to 180: $45 \left( \frac{{{25}^{x - 1}} + 7}{{{5^{x - 1}} + 1}} \right) = 180$ To isolate the fractional expression, divide both sides by 45: $\frac{{{25}^{x - 1}} + 7}{{{5^{x - 1}} + 1}} = \frac{180}{45}$ $\frac{{{25}^{x - 1}} + 7}{{{5^{x - 1}} + 1}} = 4$

Step 2.4: Solve the resulting algebraic equation using substitution.

Notice that ${25^{x - 1}}$ can be written in terms of ${5^{x - 1}}$: ${25^{x - 1}} = (5^2)^{x - 1} = (5^{x - 1})^2$ To simplify the equation, let's introduce a substitution. Let $t = {5^{x - 1}}$. Then the equation becomes: $\frac{t^2 + 7}{t + 1} = 4$ Now, solve for $t$: $t^2 + 7 = 4(t + 1)$ $t^2 + 7 = 4t + 4$ Rearrange into a standard quadratic equation form ($at^2+bt+c=0$): $t^2 - 4t + 3 = 0$ Factor the quadratic equation: We need two numbers that multiply to 3 and add up to -4. These numbers are -1 and -3. $(t - 1)(t - 3) = 0$ This gives us two possible values for $t$: $t = 1 \quad \text{or} \quad t = 3$

Step 2.5: Determine the possible values of 'x'.

Now, we substitute back $t = {5^{x - 1}}$ for each value of $t$:

• Case 1: $t=1$ ${5^{x - 1}} = 1$ Since $1 = 5^0$, we can write: ${5^{x - 1}} = {5^0}$ Equating the exponents (because the bases are the same): $x - 1 = 0$ $x = 1$

• Case 2: $t=3$ ${5^{x - 1}} = 3$ To solve for $x-1$, we take $\log_5$ on both sides: $x - 1 = \log_5 3$ $x = 1 + \log_5 3$

Step 2.6: Match with the given options.

The possible values for $x$ are $1$ and $1 + \log_5 3$. Let's check the given options: (A) 0 (B) -1 (C) 2 (D) 1

The value $x=1$ matches option (D). Self-correction/Discrepancy Note: The provided "Correct Answer" in the problem statement is (A) 0. However, our step-by-step derivation, consistent with the provided solution snippet's logic, yields $x=1$ and $x = 1 + \log_5 3$. Based on our calculations, $x=0$ is not a solution as it does not satisfy the derived equation. If $x=0$, then $t=5^{-1}=1/5$, and $\frac{(1/5)^2+7}{1/5+1} = \frac{1/25+7}{6/5} = \frac{176/25}{6/5} = \frac{88}{15} \neq 4$. Therefore, we adhere to the derived solution $x=1$.

---

3. Tips for Success and Common Mistakes

• Logarithm Properties: Always be careful when applying logarithm properties, especially when dealing with negative exponents or coefficients. Ensure the base of the logarithm matches the base of the exponential term for direct simplification.
• Identifying Terms: Pay close attention to the phrasing "increasing powers of..." to correctly identify which term is $b$ in the $(a+b)^n$ expansion. If it were "increasing powers of the first term", the role of $a$ and $b$ would be swapped in the general term formula's exponent.
• Substitution: For complex expressions involving powers of a common base (like $5^{x-1}$ and $25^{x-1}$), using a temporary variable (e.g., $t$) simplifies the algebra significantly. Remember to substitute back to find the original variable $x$.
• Quadratic Equations: Don't forget to solve quadratic equations fully, as they can yield multiple solutions for the intermediate variable ($t$), which in turn can lead to multiple solutions for $x$.

---

4. Summary and Key Takeaway

This problem is an excellent illustration of how to methodically simplify complex expressions involving logarithms and exponents, apply the Binomial Theorem to find a specific term, and then solve the resulting algebraic equation. The key steps involved recognizing and applying logarithm rules to simplify the binomial terms, correctly setting up the general term from the Binomial Theorem, performing an algebraic substitution to solve the resulting equation, and finally, interpreting the solutions for $x$. This systematic approach is crucial for tackling such multi-concept problems in mathematics.