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Step-by-step Solution
Step 1: Recognize the binomial expansion structure
We have an expression of the form
$ \bigl( A + B \bigr)^{10} $,
where
$ A = 3^{ \log_3 \sqrt{25^{x-1} + 7} } $
and
$ B = 3^{ \bigl(-\tfrac{1}{8}\bigr) \log_3 \bigl(5^{x-1} + 1\bigr) }$.
The question specifies that the ninth term, in the increasing powers of $B$, equals 180.
Step 2: Identify how the ninth term in the binomial expansion appears
In a general binomial expansion
$ (A + B)^{10} = \sum_{k=0}^{10} \binom{10}{k} A^{10-k} \, B^{k},$
the $(k+1)$th term is
$ \binom{10}{k} \, A^{10-k} \, B^{k}.$
Since we want the ninth term, set $k = 8$. So the ninth term is:
$ \binom{10}{8} \, A^{10-8} \, B^{8} = \binom{10}{8} \, A^{2} \, B^{8}.$
Step 3: Substitute the expressions for A and B
• $A = 3^{ \log_3 \sqrt{25^{x-1} + 7} } = \bigl(\sqrt{25^{x-1} + 7}\bigr)$.
This simplifies to $ \sqrt{25^{x-1} + 7} = \sqrt{ (5^{x-1})^2 + 7 }.$
• $B = 3^{ \bigl(-\tfrac{1}{8}\bigr) \log_3 (5^{x-1} + 1) } = (5^{x-1} + 1)^{-\tfrac{1}{8}}.$
Hence the ninth term becomes
$ \binom{10}{8} \bigl(\sqrt{25^{x-1} + 7}\bigr)^{2} \bigl((5^{x-1} + 1)^{-\tfrac{1}{8}}\bigr)^{8}.$
Step 4: Simplify the ninth term expression
• $ \binom{10}{8} = \binom{10}{2} = 45.$
• $\bigl(\sqrt{25^{x-1} + 7}\bigr)^{2} = 25^{x-1} + 7.$
• $\bigl((5^{x-1} + 1)^{-\tfrac{1}{8}}\bigr)^{8} = (5^{x-1} + 1)^{-1}.$
Thus, the ninth term is:
$ 45 \times \bigl(25^{x-1} + 7\bigr) \times (5^{x-1} + 1)^{-1}.$
We are given that this equals 180:
$$
45 \times \frac{25^{x-1} + 7}{5^{x-1} + 1} = 180.
$$
Step 5: Solve for x
Divide both sides by 45:
$$
\frac{25^{x-1} + 7}{5^{x-1} + 1} = 4.
$$
Let $ t = 5^{x-1}. $ Then $25^{x-1} = (5^{x-1})^2 = t^2.$ Substituting gives:
$$
\frac{t^2 + 7}{t + 1} = 4.
$$
Multiply through by $(t+1)$:
$$
t^2 + 7 = 4(t + 1).
$$
Expand and rearrange:
$$
t^2 + 7 = 4t + 4 \quad \Longrightarrow \quad t^2 - 4t + 3 = 0.
$$
Factorize:
$$
(t - 3)(t - 1) = 0.
$$
Hence $t$ can be $3$ or $1$.
If $ t = 5^{x-1} = 1, $ then:
$$
5^{x-1} = 1 \quad\Longrightarrow\quad x - 1 = 0 \quad\Longrightarrow\quad x = 1.
$$
Thus, one possible value of $x$ is $1$.
Step 6: Conclude the answer
The value of $x$ that makes the ninth term equal to 180 is
$ x = 1. $
