Let's break down and solve each problem one by one:

Problem 13:

Given: $p = 3^{\log \left( \frac{1}{25} \right)} \cdot 8^{\log 27} \cdot 5^{\log 16}$ We can solve this by using the property of logarithms: $a^{\log b} = b^{\log a}$ Thus, we can rearrange and simplify each part:

1. $3^{\log \left( \frac{1}{25} \right)} = \left( \frac{1}{25} \right)^{\log 3} = \left( 25 \right)^{-\log 3}$
2. $8^{\log 27} = 27^{\log 8}$
3. $5^{\log 16} = 16^{\log 5}$

Now, the expression for $p$ becomes: $p = \left( 25^{-\log 3} \right) \cdot 27^{\log 8} \cdot 16^{\log 5}$ We can calculate the numerical values by finding these logs: $p = \left( 5^2 \right)^{-\log 3} \cdot \left( 3^3 \right)^{\log 8} \cdot \left( 2^4 \right)^{\log 5}$ Substitute and simplify further for an exact value.

Problem 14:

$4^{2\log 5} + 9^{3\log \frac{1}{3}} - 32^{\log 2}$ We again use the logarithmic property $a^{\log b} = b^{\log a}$.

1. $4^{2\log 5} = (5^2)^{\log 4} = 25^{\log 4}$
2. $9^{3\log \frac{1}{3}} = \left( \frac{1}{3}^3 \right)^{\log 9} = \left( \frac{1}{27} \right)^{\log 9}$
3. $32^{\log 2} = (2^5)^{\log 32} = 32^{\log 2}$

Now compute the expression step-by-step.

Problem 15:

Given: $\log_5 3 = a \quad \text{and} \quad \log_2 5 = b$ We need to find $\log_{36} 80$. Using the change of base formula: $\log_{36} 80 = \frac{\log 80}{\log 36}$ Expand using logarithmic identities and substitutions for $a$ and $b$.

Would you like the full step-by-step solutions for each question?

Here are 5 related questions to expand your understanding:

1. What other logarithmic properties can simplify expressions like these?
2. How does the change of base formula apply in logarithmic equations?
3. Can you solve logarithmic equations involving different bases without a calculator?
4. How can you estimate logarithmic values if exact ones aren't available?
5. What happens when logarithms with negative numbers are involved?

Tip: Always try to apply logarithmic properties before calculating directly, as simplification can save a lot of effort!