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

Binomial Theorem — JEE Main · 2021 · 27 Jul 2021 · Shift 2 Question with Solution

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^{(- \frac{1}{8}) l o g_{3} (5^{x - 1} + 1)}}^{10}

in the increasing powers of

3^{(- \frac{1}{8}) l o g_{3} (5^{x - 1} + 1)}

is equal to 180, is :

Answer: (A)

Since

3^{l o g_{3} t} = t

for

t > 0

, write

u = 3^{l o g_{3} 2 5^{x - 1} + 7} = 2 5^{x - 1} + 7, v = 3^{- \frac{1}{8} l o g_{3} (5^{x - 1} + 1)} = (5^{x - 1} + 1)^{- 1/8}

In increasing powers of

v

, the general term of

(u + v)^{10}

(r 10) u^{10 - r} v^{r}

, so the ninth term has

r = 8

(8 10) u^{2} v^{8} = 45 \cdot \frac{2 5 ^{x - 1} + 7}{5 ^{x - 1} + 1} = 180

Hence

\frac{2 5 ^{x - 1} + 7}{5 ^{x - 1} + 1} = 4

. Putting

y = 5^{x - 1} > 0

, so that

2 5^{x - 1} = y^{2}

y^{2} + 7 = 4 (y + 1) ⟹ y^{2} - 4 y + 3 = 0 ⟹ y = 1 or y = 3

The root

y = 1

gives

5^{x - 1} = 1

, so

x = 1

, while

y = 3

gives the non-integral

x = 1 + lo g_{5} 3

. A possible value is

x = 1

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option