A possible value of x, for which the ninth term in the expansion of...

${\left\{{3}^{\log_{3}\sqrt{{25}^{x - 1} + 7}} + {3}^{\left(- \dfrac{1}{8}\right)\log_{3}\left({5}^{x - 1} + 1\right)}\right\}}^{10}$

$= {\left\{\sqrt{{25}^{x - 1} + 7} + \left({5}^{x - 1} + 1\right)\dfrac{- 1}{8}\right\}}^{10},$ using ${a}^{n{\log}_{a}x} = {x}^{n}$

Using the formula of the general term in the Binomial theorem for non negative integral index, i.e., ${T}_{r + 1} = Cr n {a}^{n - r}{b}^{r} in${\left(a + b\right)}^{n}, we get, ${T}_{9} = {T}_{8 + 1} = C8 10 \left({25}^{\left(x - 1\right)} + 7\right) \times {\left({5}^{\left(x - 1\right)} + 1\right)}^{- 1} = 180$

$⇒ \dfrac{{25}^{x - 1} + 7}{{5}^{\left(x - 1\right)} + 1} = 4$

$⇒ \dfrac{{t}^{2} + 7}{t + 1} = 4,$ taking $t = {5}^{x - 1}$

$⇒ {t}^{2} + 7 = 4\left(t + 1\right)$

$⇒ {t}^{2} - 4t + 3 = 0$

$⇒ \left(t - 3\right)\left(t - 1\right) = 0$

$⇒ t = 1, 3 ⇒ {5}^{x - 1} = 1 ⇒ {5}^{x - 1} = {5}^{0}$

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

$⇒ x = 1$