https://r2.erweima.ai/mathbot/image/1726536735520fuqa9at4.jpg

The mathematical expression in the image is:

\[
\left( \log 10 (100 + \sqrt[3]{8}) \times 5^3 \right) - \left( \log 9 (81 \times 5) \right)
\]

Let's break it down:

1. The first term:
   \[
   \log 10 \left( 100 + \sqrt[3]{8} \right) \times 5^3
   \]
   where \( \log 10 \) is the logarithm base 10, \( \sqrt[3]{8} \) is the cube root of 8 (which equals 2), and \( 5^3 = 125 \).

   - So, this becomes:
     \[
     \log 10 (100 + 2) \times 125 = \log 10 (102) \times 125
     \]
     Using the logarithmic value \( \log_{10} 102 \approx 2.0086 \), we get:
     \[
     2.0086 \times 125 \approx 251.075
     \]

2. The second term:
   \[
   \log 9 (81 \times 5)
   \]
   where \( \log 9 \) is the logarithm base 9, and \( 81 \times 5 = 405 \).

   - Using the change of base formula for logarithms:
     \[
     \log_9 405 = \frac{\log_{10} 405}{\log_{10} 9}
     \]
     where \( \log_{10} 405 \approx 2.6075 \) and \( \log_{10} 9 \approx 0.9542 \), we get:
     \[
     \frac{2.6075}{0.9542} \approx 2.732
     \]

Thus, the full expression becomes:
\[
251.075 - 2.732 \approx 248.343
\]

The final result is approximately \( 248.343 \).

Would you like a deeper explanation or clarification on any part? Here are 5 related questions for further exploration:

1. How does the change of base formula for logarithms work?
2. Can you explain the cube root and how it is calculated?
3. Why is the base of logarithms important in calculations?
4. What are some real-life applications of logarithmic functions?
5. How can logarithmic and exponential functions be used in compound interest?

**Tip:** When using logarithms with different bases, the change of base formula is an essential tool for simplifying complex expressions.

(log10(100 + ³√8) * 5³) - (log9(81 * 5))

This problem involves solving a complex logarithmic expression, combining the change of base formula and properties of logarithms. Step-by-step breakdown provided for cube roots, powers, and base 9 logarithms.

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