First slide

Binomial theorem for positive integral Index

Question

The value of x, for which the 6th term in the expansion of $\{2^{\log_{2} \sqrt{(9^{x - 1} + 7)}} + \frac{1}{2^{(1 / 5) \log_{2} (3^{x - 1} + 1)}}\}$ is 84,is equal to

Moderate

A

4
B

3
C

2
D

5

Solution

We have, ${[2^{\log_{2} \sqrt{9^{x - 1} + 7}} + \frac{1}{2^{(\frac{1}{5}) \log_{2} (3^{x - 1} + 1)}}]}^{7}$
$= {[\sqrt{9^{x - 1} + 7} + \frac{1}{{(3^{x - 1} + 1)}^{1 / 5}}]}^{7}$
$∴ T_{6} =^{7} C_{5} {(\sqrt{9^{x - 1} + 7})}^{7 - 5} {[\frac{1}{{(3^{x - 1} + 1)}^{1 / 5}}]}^{5}$
$=^{7} C_{5} (9^{x - 1} + 7) \frac{1}{(3^{x - 1} + 1)}$
$\Rightarrow 84 =^{7} C_{5} \frac{(9^{x - 1} + 7)}{(3^{x - 1} + 1)}$
$\Rightarrow 9^{x - 1} + 7 = 4 (3^{x - 1} + 1)$
$\Rightarrow \frac{3^{2 x}}{9} + 7 = 4 (\frac{3^{x}}{3} + 1)$
$\Rightarrow y^{2} - 12 y + 27 = 0$
$\Rightarrow y^{2} - 12 y + 27 = 0$ (Put y= 3)
$\Rightarrow (y - 3) (y - 9) = 0$
$\Rightarrow y = 3, 9 \Rightarrow 3^{x} = 3, 9 \Rightarrow x = 1, 2$

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