Question:
The sixth term in the expansion of
$\left[2^{\log _{2} \sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5} \log _{2}\left(3^{x-1}+1\right)}}\right]$ is 84.
Then the number of values of $x$ is
The sixth term in the expansion of
$\left[2^{\log _{2} \sqrt{9^{x-1}+7}}+\frac{1}{2^{\frac{1}{5} \log _{2}\left(3^{x-1}+1\right)}}\right]$ is 84.
Then the number of values of $x$ is
Updated On: Jul 7, 2022
- $0$
- $1$
- $2$
- $3$
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The Correct Option is C
Solution and Explanation
$2^{log_2^{\sqrt{9^{x-1}+7}}} = \sqrt{9^{x-1}+7}$
$2^{\frac{1}{2}log_{2}\left(3^{x-1}+1\right)}$
$= \left(3^{x-1}+1\right)^{1/5}$
Also, the sixth term in the expansion is $84$
$\therefore \,^{7}C_{5}\left( \sqrt{9^{x-1}+7}\right)^{2}\cdot\frac{1}{3^{x-1}+1} = 84$
$\Rightarrow \frac{9^{x-1}+7}{3^{x-1}+1} = \frac{84}{21}$
$\Rightarrow \left( 3^{x-1}\right)^{2}- 4 \left( 3^{x-1}\right) + 3 = 0$
$\Rightarrow \left(3^{x-1}-1\right)\left(3^{x-1}-3\right) = 0$
$\therefore 3^{x-1} = 1$ or $3$
$\Rightarrow x = 1$ or $2$.
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Notes on binomial expansion formula
Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .