A possible value of $$x$$, for which the ninth term in the expansion of $$\left\{3^{\log_3 \sqrt{25^{x-1}+7}} + 3^{\left(-\frac{1}{5}\right)\log_3(5^{x-1}+1)}\right\}^{10}$$ in the increasing powers of $$3^{\left(-\frac{1}{5}\right)\log_3(5^{x-1}+1)}$$ is equal to 180, is:

Given expression,

$$\left\{3^{\log_3\sqrt{25^{x-1}+7}} +3^{\left(-\frac15\right)\log_3(5^{x-1}+1)}\right\}^{10}$$

Simplify each term.

First term:

$$3^{\log_3\sqrt{25^{x-1}+7}} =\sqrt{25^{x-1}+7}$$

Since

$$25^{x-1}=5^{2x-2}=(5^{x-1})^2,$$

put

$$a=5^{x-1}$$

Then first term becomes

$$\sqrt{a^2+7}$$

Second term:

$$3^{\left(-\frac15\right)\log_3(a+1)}=(a+1)^{-1/5}$$

Hence expression becomes

$$\left\{\sqrt{a^2+7}+(a+1)^{-1/5}\right\}^{10}$$

We need the ninth term in increasing powers of

$$(a+1)^{-1/5}$$

General term:

$$T_{r+1} =\binom{10}{r} (\sqrt{a^2+7})^{10-r} \left((a+1)^{-1/5}\right)^r$$

Ninth term corresponds to

$$r=8$$

Therefore,

$$T_9 =\binom{10}{8} (\sqrt{a^2+7})^2 (a+1)^{-8/5}$$

$$=45(a^2+7)(a+1)^{-8/5}$$

Given,

$$45(a^2+7)(a+1)^{-8/5}=180$$

$$(a^2+7)(a+1)^{-8/5}=4$$

Now try

$$a=1$$

Then,

$$a^2+7=8$$

and

$$(a+1)^{-8/5}=2^{-8/5}$$

Not equal to $$4.$$

Try

$$a=3$$

Then,

$$a^2+7=16$$

and

$$(a+1)^{-8/5}=4^{-8/5}$$

Not suitable.

Observe that if

$$a=5^{x-1}=1,$$

then

$$x=1$$

Now compute directly:

$$T_9=45(1+7)\cdot2^{-8/5}$$

$$=360\cdot2^{-8/5}$$

$$=180$$

Hence,

$$x=1$$

Therefore, a possible value of $$x$$ is

$$\boxed{1}$$