The problem is asking to use the logarithmic identity $a^{\frac{\log b}{\log a}} = b$ to solve the expressions provided.

Here are the expressions and the options shown in the image:

1. Expression (a): $100^{\frac{\log 8}{\log 5}}$
2. Expression (b): 32
3. Expression (c): $25^{\frac{\log 8}{\log 5}} + 64^{\frac{\log 8}{\log 5}}$
4. Expression (d): $4^{\frac{\log 7}{\log 2}} - 8^{\frac{\log 7}{\log 3}} + 3$

Let's analyze each one:

1. (a) $100^{\frac{\log 8}{\log 5}}$: Using the property $a^{\frac{\log b}{\log a}} = b$, this expression simplifies to 8.

2. (b) 32: This is a simple number, no calculations required.

3. (c) $25^{\frac{\log 8}{\log 5}} + 64^{\frac{\log 8}{\log 5}}$: Both terms follow the same property:

   • $25^{\frac{\log 8}{\log 5}} = 8$
   • $64^{\frac{\log 8}{\log 5}} = 8$

   Thus, $8 + 8 = 16$.

4. (d) $4^{\frac{\log 7}{\log 2}} - 8^{\frac{\log 7}{\log 3}} + 3$: This one is more complex, but using the logarithmic identity for each term:

   • $4^{\frac{\log 7}{\log 2}} = 7$
   • $8^{\frac{\log 7}{\log 3}} = 7$

   So, the expression simplifies to $7 - 7 + 3 = 3$.

Final answers:

• a) 8
• b) 32
• c) 16
• d) 3

Would you like any further explanation on how these simplifications were done?

Here are five related questions to deepen your understanding:

1. What is the logarithmic identity $a^{\frac{\log b}{\log a}} = b$ used for?
2. How can logarithmic identities help simplify expressions involving exponents?
3. What other logarithmic properties can be used in similar problems?
4. How would the calculations change if the base of the logarithms was different?
5. Can you solve $x^{\frac{\log y}{\log z}}$ for various values of $x, y, z$?

Tip: When working with logarithmic identities, remember that simplifying the base often helps reduce complex expressions quickly.