The expression shown in the image is:

${}^{3} \log 5 \times {}^{25} \log 27$

This is written using logarithms with different bases, and we will break it down.

1. Rewrite the logarithms with their respective bases:
   • ${}^{3} \log 5$ means the logarithm of 5 with base 3, i.e., $\log_3 5$.
   • ${}^{25} \log 27$ means the logarithm of 27 with base 25, i.e., $\log_{25} 27$.

Thus, the expression becomes:

$\log_3 5 \times \log_{25} 27$

2. Change of base formula: We can simplify these logarithms using the change of base formula, which states:

   $\log_b a = \frac{\log a}{\log b}$

   Applying this:

   \quad \text{and} \quad \log_{25} 27 = \frac{\log 27}{\log 25}$$

So the expression now becomes:

$\frac{\log 5}{\log 3} \times \frac{\log 27}{\log 25}$

3. Simplification:
   • $\log 27$ can be written as $\log (3^3) = 3 \log 3$ by using logarithmic properties.
   • $\log 25$ can be written as $\log (5^2) = 2 \log 5$.

Thus, we now have:

$\frac{\log 5}{\log 3} \times \frac{3 \log 3}{2 \log 5}$

4. Cancel terms:
   • $\log 5$ cancels with $\log 5$.
   • $\log 3$ cancels with $\log 3$.

Now the expression simplifies to:

$\frac{3}{2}$

So, the final answer is:

$\boxed{\frac{3}{2}}$

Would you like more details or have any further questions?

Here are 5 related questions to deepen your understanding:

1. What is the change of base formula and how is it derived?
2. How do logarithmic properties, such as $\log(a^b) = b \log a$, simplify complex expressions?
3. Can the logarithms of other bases be converted into common logarithms (base 10)?
4. How can logarithmic identities be used in solving exponential equations?
5. What are the practical applications of logarithmic calculations in real-world problems?

Tip: Always simplify expressions using logarithmic properties before applying numerical values to make the calculation easier!