The given expression is:

$25^{\left(\frac{1}{2} + \log_1^5 27 + \log_{25} 81 \right)}$

Let's break it down step by step for a Class 9 level.

Step 1: Simplify the logarithms

1. First term: $\frac{1}{2}$ remains as is.

2. Second term: $\log_1^5 27 = \log_{10} 27$ Since the logarithm with a subscript $1^5$ would actually be $\log_{10} 27$, which can be estimated but isn't necessary for now.

3. Third term: $\log_{25} 81 = \frac{\log 81}{\log 25}$

Check th