The given expression to simplify is:

$\log_{15}^{2} 3 + \frac{\log_{15} 45}{\log_{5} 15}$

Step 1: Simplify the first term, $\log_{15}^{2} 3$.

This term can be written as:

$\left(\log_{15} 3\right)^2$

Step 2: Simplify the second term, $\frac{\log_{15} 45}{\log_{5} 15}$.

First, let's break down $\log_{15} 45$ using the change of base formula:

$\log_{15} 45 = \frac{\log 45}{\log 15}$

Similarly, express $\log_{5} 15$ using the change of base formula:

$\log_{5} 15 = \frac{\log 15}{\log 5}$

Now, substitute these into the expression:

$\frac{\log_{15} 45}{\log_{5} 15} = \frac{\frac{\log 45}{\log 15}}{\frac{\log 15}{\log 5}} = \frac{\log 45}{\log 15} \times \frac{\log 5}{\log 15} = \frac{\log 45 \cdot \log 5}{(\log 15)^2}$

Step 3: Break down $\log 45$ and simplify.

Notice that:

$\log 45 = \log (3 \times 15) = \log 3 + \log 15$

So, the expression becomes:

$\frac{(\log 3 + \log 15) \cdot \log 5}{(\log 15)^2}$

Distribute $\log 5$ in the numerator:

$\frac{\log 3 \cdot \log 5 + (\log 15) \cdot \log 5}{(\log 15)^2}$

This can be separated into two fractions:

$\frac{\log 3 \cdot \log 5}{(\log 15)^2} + \frac{(\log 15) \cdot \log 5}{(\log 15)^2}$

The second fraction simplifies to:

$\frac{\log 5}{\log 15}$

So the original expression becomes:

$\left(\log_{15} 3\right)^2 + \frac{\log 3 \cdot \log 5}{(\log 15)^2} + \frac{\log 5}{\log 15}$

Step 4: Express everything in terms of logarithms to base 15.

Notice that:

$\log_{15} 3 = \frac{\log 3}{\log 15} \quad \text{and} \quad \log_{15} 5 = \frac{\log 5}{\log 15}$

Substituting back, the expression becomes:

$\left(\log_{15} 3\right)^2 + \log_{15} 3 \cdot \log_{15} 5 + \log_{15} 5$

Final Answer:

The simplified expression is:

$\left(\log_{15} 3 + \log_{15} 5\right)^2 = \left(\log_{15} 15\right)^2 = 1$

So, the original expression simplifies to $1$.

Would you like more details or have any further questions?

Related Questions:

1. How would the expression change if the base was 10 instead of 15?
2. Can the expression be generalized for any base $b$?
3. What is the value of $\log_{15} 3$ and $\log_{5} 15$ approximately?
4. How would the expression change if $\log_{15} 45$ was replaced with $\log_{15} 27$?
5. Can this expression be simplified using properties of exponents?

Tip:

Always check if logarithmic expressions can be simplified by combining or expanding them using properties like $\log(ab) = \log a + \log b$.