68) The value of x, for which the 6th term in the expansion of s 2log29x-1+7+1215log23x-1+17 is 84 is equal - Maths - Binomial Theorem

Question

68) The value of x, for which the 6 t h term in the expansion of s
2 log 2 9 x - 1 + 7 + 1 2 1 5 log 2 3 x - 1 + 1 7 is 84 is equal
to
(a) 4 , 3
(b) 0 , 3
(c) 0 , 2
(d) 1 ,2

Rahul Raj · Accepted Answer

dear student

7th term=84C57(2log29x-1+7)2(2-15log2(3x-1+1))5=847!2!5!22*12*log2(9x-1+7)25*-15*log2(3x-1+1)=8421×2log2(9x-1+7)2-1*log2(3x-1+1)=842log2(9x-1+7)-log2(3x-1+1)=42log2(9x-1+7)(3x-1+1)=4=22log2(9x-1+7)(3x-1+1)=2(9x-1+7)(3x-1+1)=4a2+7a+1=4 where a=3x-1a2-4a+3=0a=1 or a=33x-1=1or 3x=1,2

regards

68) The value of x, for which the 6 t h term in the expansion of s 2 log 2 9 x - 1 + 7 + 1 2 1 5 log 2 3 x - 1 + 1 7 is 84 is equal to (a) 4 , 3 (b) 0 , 3 (c) 0 , 2 (d) 1 ,2

68) The value of x, for which the $6^{t h}$ term in the expansion of s $\{2^{\log_{2} \sqrt{(9^{x - 1} + 7)}} + \frac{1}{2^{(\frac{1}{5}) \log_{2} (3^{x - 1} + 1)}}\}$ $^{7}$ is 84 is equal to
(a) 4 , 3
(b) 0 , 3
(c) 0 , 2
(d) 1 ,2