Question

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

• A. 0
• B. 1
• C. -1
• D. 2

Hint:

Simplify the terms within the binomial expansion first using logarithm properties like $a^{\log_a b} = b$. Then, write out the specific term required by the question. Often, this leads to an equation that can be solved by substitution.

Answer 1

The Correct Option is B

Let's simplify the terms in the binomial expansion.
Let $A = 3^{\log_3\sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$.
Let $B = 3^{(-\frac{1}{8})\log_3(5^{x-1}+1)} = (5^{x-1}+1)^{-1/8}$.
The expansion is $(A+B)^{10}$.
The general term is $T_{r+1} = \binom{10}{r} A^{10-r} B^r$.
The ninth term corresponds to $r=8$.
$T_9 = \binom{10}{8} A^{10-8} B^8 = \binom{10}{2} A^2 B^8$.
We are given that $T_9 = 180$.
$\binom{10}{2} = \frac{10 \times 9}{2} = 45$.
$45 A^2 B^8 = 180$.
$A^2 B^8 = \frac{180}{45} = 4$.
Now substitute the expressions for A and B.
$A^2 = (\sqrt{25^{x-1}+7})^2 = 25^{x-1}+7 = (5^2)^{x-1}+7 = 5^{2x-2}+7$.
$B^8 = ((5^{x-1}+1)^{-1/8})^8 = (5^{x-1}+1)^{-1} = \frac{1}{5^{x-1}+1}$.
Substitute these into the equation $A^2 B^8 = 4$:
$(5^{2x-2}+7) \cdot \frac{1}{5^{x-1}+1} = 4$.
$5^{2x-2}+7 = 4(5^{x-1}+1)$.
$(5^{x-1})^2 + 7 = 4(5^{x-1}) + 4$.
Let $y = 5^{x-1}$. The equation becomes a quadratic in y:
$y^2 + 7 = 4y + 4$.
$y^2 - 4y + 3 = 0$.
Factor the quadratic:
$(y-1)(y-3) = 0$.
So, $y=1$ or $y=3$.
Case 1: $y = 1$.
$5^{x-1} = 1 = 5^0$.
$x-1 = 0 \implies x=1$.
Case 2: $y = 3$.
$5^{x-1} = 3$.
$x-1 = \log_5(3) \implies x = 1 + \log_5(3)$.
The options are integers. The value $x=1$ is a possible value and corresponds to option (B).