Answer :

To evaluate the expression [tex]2 \log_4 64 + \log_3 (9\sqrt{3}) - 3 \log_5 2[/tex], let's break it down into smaller parts and simplify each part step-by-step.


  1. Evaluate [tex]2 \log_4 64[/tex]:


    • First, recall that [tex]\log_b a = c[/tex] means [tex]b^c = a[/tex].

    • So [tex]\log_4 64[/tex] asks: "What power must 4 be raised to, to get 64?"

    • We can express 64 as a power of 4: 64 = [tex]4^3[/tex]. Thus, [tex]\log_4 64 = 3[/tex].

    • Multiply by 2: [tex]2 \times 3 = 6[/tex].



  2. Evaluate [tex]\log_3 (9\sqrt{3})[/tex]:


    • First, let's express 9 and [tex]\sqrt{3}[/tex] using base 3: 9 = [tex]3^2[/tex] and [tex]\sqrt{3} = 3^{0.5}[/tex].

    • Therefore, [tex]9\sqrt{3} = 3^2 \times 3^{0.5} = 3^{2.5}[/tex].

    • So, [tex]\log_3 (3^{2.5}) = 2.5[/tex].



  3. Evaluate [tex]3 \log_5 2[/tex]:


    • Let's find [tex]\log_5 2[/tex]. This requires more work as it's not a straightforward exponentiation.

    • However, typically in high school mathematics, you might use the change of base formula:
      [tex]\log_5 2 = \frac{\log_{10} 2}{\log_{10} 5}[/tex].

    • For simplicity, let's assume a calculator is used:

      • [tex]\log_{10} 2 \approx 0.3010[/tex]

      • [tex]\log_{10} 5 \approx 0.69897[/tex]

      • Therefore, [tex]\log_5 2 \approx \frac{0.3010}{0.69897} \approx 0.4306[/tex].



    • Multiply by 3: [tex]3 \times 0.4306 = 1.2918[/tex] (rounded).



  4. Combine all parts:


    • [tex]2 \log_4 64 + \log_3 (9\sqrt{3}) - 3 \log_5 2 = 6 + 2.5 - 1.2918[/tex].

    • Add and subtract the values: [tex]6 + 2.5 - 1.2918 = 7.2082[/tex].




Therefore, the evaluated expression is approximately [tex]7.2082[/tex].