A possible value of x, for which the ninth term in the expansion of {3^{log₃ √(25^{x-1}+7)} + 3^{(-1

Q: Why can't we just expand the whole thing? It is only power 10.

Technically, you could, but that is a path to madness! In JEE Advanced, we look for elegance. Expanding a binomial with logarithmic terms is a trap designed to waste your time. Always simplify the base terms first. The beauty of this problem lies in the simplification, not the brute force.

Q: How did you know to use the substitution $t = 5^{x-1}$?

Look at the terms we derived. We had $25^{x-1}$ and $5^{x-1}$. Since $25 = 5^2$, the expression $25^{x-1}$ is just $(5^{x-1})^2$. Whenever you see a variable and its square, substitution is your best friend. It turns a scary exponential equation into a simple quadratic one.

Q: I got two values for $t$, but only one answer for $x$. Did I do something wrong?

Not at all! You found two mathematically valid roots for the quadratic. However, in a multiple-choice question, we must check which of those roots leads to an answer present in the options. Always keep an eye on the target. You solved it perfectly.

Question

A possible value of $x$, for which the ninth term in the expansion of $\{3^{\log_3 \sqrt{25^{x-1}+7}} + 3^{(-1/8) \log_3(5^{x-1}+1)}\}^{10}$ in the increasing powers of $3^{(-1/8) \log_3(5^{x-1}+1)}$ is equal to 180, is :

Options:

(A) 0

(B) $-1$

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression provided is:
$$\left\{ 3^{\log_3 \sqrt{25^{x-1}+7}} + 3^{(-1/8) \log_3(5^{x-1}+1)} ight\}^{10}$$

Our objective is to find the value of $x$ such that the ninth term of this binomial expansion equals $180$. We begin by simplifying the base terms.

### The Art of Simplification

Consider the first term, $a = 3^{\log_3 \sqrt{25^{x-1}+7}}$. Using the fundamental logarithmic identity $k^{\log_k X} = X$, the base $3$ and the $\log_3$ cancel out:
$$a = \sqrt{25^{x-1}+7}$$

Now, consider the second term, $b = 3^{(-1/8) \log_3(5^{x-1}+1)}$. Applying the power rule $n \log X = \log X^n$, we move the coefficient $-1/8$ inside the logarithm:
$$b = 3^{\log_3 (5^{x-1}+1)^{-1/8}} = (5^{x-1}+1)^{-1/8}$$

### The Binomial Dance

With $a$ and $b$ simplified, the expression is $(a+b)^{10}$. The general term for a binomial expansion is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

For the ninth term, we set $r=8$ and $n=10$:
$$T_9 = \binom{10}{8} a^{10-8} b^8 = \binom{10}{2} a^2 b^8$$

Calculating the binomial coefficient, we find $\binom{10}{2} = \frac{10 	imes 9}{2 	imes 1} = 45$. Substituting the expressions for $a^2$ and $b^8$:
$$a^2 = 25^{x-1}+7$$
$$b^8 = \left( (5^{x-1}+1)^{-1/8} ight)^8 = (5^{x-1}+1)^{-1} = \frac{1}{5^{x-1}+1}$$

### The Quadratic Reveal

Setting the ninth term equal to $180$, we have:
$$45 \cdot \frac{25^{x-1}+7}{5^{x-1}+1} = 180$$

Dividing both sides by $45$ yields:
$$\frac{25^{x-1}+7}{5^{x-1}+1} = 4$$

Let $t = 5^{x-1}$. Since $25^{x-1} = (5^{x-1})^2 = t^2$, the equation becomes:
$$\frac{t^2+7}{t+1} = 4$$

Cross-multiplying results in the quadratic equation:
$$t^2 + 7 = 4t + 4 \Rightarrow t^2 - 4t + 3 = 0$$

### Final Calculation

Factoring the quadratic equation $(t-1)(t-3) = 0$, we obtain two possible values for $t$:
$$t = 1 \quad 	ext{or} \quad t = 3$$

Substituting back $t = 5^{x-1}$:
1. $5^{x-1} = 1 \Rightarrow x-1 = 0 \Rightarrow \mathbf{x=1}$
2. $5^{x-1} = 3 \Rightarrow x-1 = \log_5 3 \Rightarrow \mathbf{x = 1 + \log_5 3}$

Both values are mathematically valid solutions for the given condition.

Animated Solution for Mathematics - Binomial Theorem: A possible value of $x$ , for which the ninth term in the expansion of ${3^{l o g_{3} 2 5^{x - 1} + 7} + 3^{(- 1/8) l o g_{3} (5^{x - 1} + 1)}}^{10}$ in the increasing powers of $3^{(- 1/8) l o g_{3} (5^{x - 1} + 1)}$ is equal to 180, is :

Select Answer:

Visualized Solution

Analyzing the Expression $(a + b)^{10}$

Simplifying the First Term $a$

Simplifying the Second Term $b$

The General Term Formula $T_{r + 1}$

Setting up the 9th Term $T_{9}$

Substituting $a$ and $b$ into $T_{9}$

Equating $T_{9}$ to $180$

Preparing for Substitution $t = 5^{x - 1}$

Substitution to Quadratic Form

Solving the Quadratic Equation for $t$

Back-substitution to find $x$

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Art of Simplification

The Binomial Dance

The Quadratic Reveal

Final Calculation

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Select Answer:

Visualized Solution

Analyzing the Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a+b)10

Simplifying the First Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Simplifying the Second Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The General Term Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tr+1​

Setting up the 9th Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T9​

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T9​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 180

Preparing for Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=5x−1

Substitution to Quadratic Form

Solving the Quadratic Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Back-substitution to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x