Log!

$\text {Let } a = 5^{x-1}$

Then:

$3 ^ { \log _3 \sqrt { 25^ {x -1} + 7 } } = \sqrt { 25^ {x -1} + 7 } = \sqrt { a^2 + 7 }$

$3 ^ { - \frac {1}{8} \log _3 { 5^ {x -1} + 1 } } = (5^ {x -1} + 1 ) ^ { - \frac {1}{8} } = \frac {1}{ \sqrt [8] { 5a + 1 } }$

Now, the 9th term of the binomial expansion of our expression:

${ 10 \choose 8 } ( \sqrt { a^2 + 7 } )^2 ( \frac {1}{ \sqrt [8] { 5a + 1 } } )^8 = 180$

$45 \frac {a^2 + 7}{a + 1} = 180$

$\frac {a^2 + 7}{a + 1} = 4$

$a^2 + 7 = 4a + 4$

$a^2 - 4a + 3 = 0$

$(a -3)(a-1) = 0$

$a = 3 \text { or } a = 1$

If a =3:

$5^{x - 1} = 3$

$x - 1 = \log _5 3$

$x = 1.6826 \text { (4 d. p.) } \Rightarrow \text { Not an integer, therefore not a solution. }$

If a =1:

$5^{x - 1} = 1$

$x - 1 = \log _5 1 = 0$

$x = \boxed {1}$