A possible value of x,for which the ninth ter | vedclass

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(B) Let $a = 3^{\log _{3} \sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$ and $b = 3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)} = (5^{x-1}+1)

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$1$

Vedclass · Answer

(B) Let $a = 3^{\log _{3} \sqrt{25^{x-1}+7}} = \sqrt{25^{x-1}+7}$ and $b = 3^{\left(-\frac{1}{8}ight) \log _{3}\left(5^{x-1}+1ight)} = (5^{x-1}+1)^{-1/8}$.The expansion is $(a+b)^{10}$. The ninth term $T_9$ is given by $T_9 = {}^{10}C_8 a^{10-8} b^8 = {}^{10}C_8 a^2 b^8$.Substituting the values: ${}^{10}C_8 = 45$,$a^2 = 25^{x-1}+7$,and $b^8 = (5^{x-1}+1)^{-1}$.So,$45 	imes \frac{25^{x-1}+7}{5^{x-1}+1} = 180$.Dividing by $45$,we get $\frac{25^{x-1}+7}{5^{x-1}+1} = 4$.Let $t = 5^{x-1}$. Then $\frac{t^2+7}{t+1} = 4$.$t^2+7 = 4t+4 \Rightarrow t^2-4t+3 = 0$.$(t-1)(t-3) = 0$,so $t=1$ or $t=3$.If $5^{x-1} = 1$,then $x-1 = 0 \Rightarrow x = 1$.If $5^{x-1} = 3$,then $x-1 = \log_5 3 \Rightarrow x = 1 + \log_5 3$.

$A$ 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:

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