Question

The ninth term in the expansion of

$\left\{ 3^{\log_{3}\sqrt{25^{x - 1} + 7}} + 3^{- 1/8\log_{3}(5^{x - 1} + 1)} \right\}^{10}$ is equal to 180, then x is

• A. 2
• B. 1
• C. 4
• D. None of these

Answer 1

Answer: (B)

1

Answer 2

We have,

$\left\{ 3^{\log_{3}\sqrt{25^{x - 1} + 7}} + 3^{- 1/8\log_{3}(5^{x - 1} + 1)} \right\}^{10}$

= $\left\lbrack \sqrt{25^{x - 1} + 7} + (5^{x - 1} + 1)^{- 1/8} \right\rbrack^{10}$

{Q $a^{\log_{a}N}$= N}

Here, T₉ = 180.

Ž ¹⁰C₈ $\left\{ \sqrt{25^{x - 1} + 7} \right\}^{10 - 8}${(5^x–1 + 1)^–1/8}⁸ = 180

Ž ¹⁰C₈(25^x–1 + 7) (5^x–1 + 1)^–1 = 180

Ž 45$\frac{(25^{x - 1} + 7)}{5^{x - 1} + 1}$ = 180

Ž $\frac{25^{x - 1} + 7}{5^{x - 1} + 1}$ = 4

Ž $\frac{y^{2} + 7}{y + 1}$= 4, where y = 5^x–1

Ž y² – 4y + 3 = 0

Ž y = 3, 1

Ž 5^x–1 = 3 or 5^x–1 = 1

Ž 5^x = 15 or 5^x = 5

Ž x = log₅15 or x = 1

Hence (2) is correct answer.