A possible value of x, for which the ninth term in the expansion of 3^ log_3√25^x-1+7+3^(- 1 8) log_3(5^x-1+1)^10 in the increasing powers of 3^(- 1…

$${{{3}^{{\mathrm{log}}_{3}\sqrt{{25}^{x-1}+7}}+{3}^{(-\frac{1}{8}){\mathrm{log}}_{3}({5}^{x-1}+1)}}}^{10}$$3log3​25x−1+7​+3(−81​)log3​(5x−1+1)10

$={{\sqrt{{25}^{x-1}+7}+{({5}^{x-1}+1)}^{\frac{-1}{8}}}}^{10},$=25x−1+7​+(5x−1+1)8−1​10, using ${a}^{n{\mathrm{log}}_{a}x}={x}^{n}$anloga​x=xn

Using the formula of the general term in the Binomial theorem for non negative integral index, i.e., ${T}_{r+1}=Crn{a}^{n-r}{b}^{r}$Tr+1​=Crnan−rbr in ${(a+b)}^{n},$(a+b)n, we get, ${T}_{9}={T}_{8+1}=C810({25}^{(x-1)}+7)\times {({5}^{(x-1)}+1)}^{-1}=180$T9​=T8+1​=C810(25(x−1)+7)×(5(x−1)+1)−1=180

$$\Rightarrow \frac{{25}^{x-1}+7}{{5}^{(x-1)}+1}=4$$⇒5(x−1)+125x−1+7​=4

$\Rightarrow \frac{{t}^{2}+7}{t+1}=4,$⇒t+1t2+7​=4, taking $t={5}^{x-1}$t=5x−1

$$\Rightarrow {t}^{2}+7=4(t+1)$$⇒t2+7=4(t+1)

$$\Rightarrow {t}^{2}-4t+3=0$$⇒t2−4t+3=0

$$\Rightarrow (t-3)(t-1)=0$$⇒(t−3)(t−1)=0

$$\Rightarrow t=1,3\Rightarrow {5}^{x-1}=1\Rightarrow {5}^{x-1}={5}^{0}$$⇒t=1,3⇒5x−1=1⇒5x−1=50

$\Rightarrow x-1=0$⇒x−1=0 (one of the possible value).

$$\Rightarrow x=1$$⇒x=1