To solve this expression, let’s break it down step-by-step:

${}^5 \log 16 + {}^2 \log 25 - {}^3 \log \frac{27}{81}$

Each term uses a different logarithmic base, so let's handle each term individually.

Step 1: Simplify ${}^5 \log 16$

This term represents the logarithm of 16 in base 5: ${}^5 \log 16 = \frac{\log 16}{\log 5}$ Rewrite 16 as $2^4$: ${}^5 \log 16 = \frac{\log (2^4)}{\log 5} = \frac{4 \log 2}{\log 5}$

Step 2: Simplify ${}^2 \log 25$

This term represents the logarithm of 25 in base 2: ${}^2 \log 25 = \frac{\log 25}{\log 2}$ Rewrite 25 as $5^2$: ${}^2 \log 25 = \frac{\log (5^2)}{\log 2} = \frac{2 \log 5}{\log 2}$

Step 3: Simplify ${}^3 \log \frac{27}{81}$

This term represents the logarithm of $\frac{27}{81}$ in base 3: ${}^3 \log \frac{27}{81} = \frac{\log \frac{27}{81}}{\log 3}$ Rewrite $\frac{27}{81}$ as $3^3 / 3^4 = 3^{-1}$: ${}^3 \log \frac{27}{81} = \frac{\log (3^{-1})}{\log 3} = \frac{-\log 3}{\log 3} = -1$

Step 4: Combine All Terms

Now we substitute each simplified term back into the original expression: $\frac{4 \log 2}{\log 5} + \frac{2 \log 5}{\log 2} - 1$

This expression is the simplified form of the given problem.

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Related Questions

1. How do you convert a logarithm from one base to another?
2. What are the common logarithmic properties and how are they used?
3. How do you simplify logarithmic expressions involving fractions?
4. What is the change of base formula for logarithms?
5. How can we simplify logarithms with exponents in the argument?

Tip:

Remember that changing the base of a logarithm can often make complex logarithmic expressions much easier to work with.