If $$ x $$ is a positive real number such that $$ 4\log_{10}x + 4\log_{100}x + 8\log_{1000}x = 13 $$, then the greatest integer not exceeding $$ x $$ is:

Question

Accepted Answer

The problem involves solving a logarithmic equation to find the value of $$ x $$ and then determining the greatest integer not exceeding $$ x $$. Here's the step-by-step solution:

**Given Equation:**

$$
   4\log_{10}x + 4\log_{100}x + 8\log_{1000}x = 13
   $$

**Rewrite in Terms of $$ \log_{10}x $$:**

- Recall that:
     $$
     \log_{100}x = \frac{\log_{10}x}{2}, \quad \log_{1000}x = \frac{\log_{10}x}{3}.
     $$
   - Substituting these into the equation:
     $$
     4\log_{10}x + 4\left(\frac{\log_{10}x}{2}\right) + 8\left(\frac{\log_{10}x}{3}\right) = 13
     $$

**Simplify the Equation:**

- Combine terms:
     $$
     4\log_{10}x + 2\log_{10}x + \frac{8}{3}\log_{10}x = 13
     $$
     - The coefficient of $$ \log_{10}x $$ is:
       $$
       4 + 2 + \frac{8}{3} = \frac{26}{3}.
       $$

- Thus, the equation becomes:
     $$
     \frac{26}{3}\log_{10}x = 13
     $$

**Solve for $$ \log_{10}x $$:**

- Divide through by $$ \frac{26}{3} $$:
     $$
     \log_{10}x = \frac{13}{\frac{26}{3}} = \frac{13 \cdot 3}{26} = \frac{39}{26} = \frac{3}{2}.
     $$

**Solve for $$ x $$:**

- Using the property of logarithms:
     $$
     x = 10^{\log_{10}x} = 10^{\frac{3}{2}}.
     $$
   - Simplify:
     $$
     x = 10^{1.5} = \sqrt{1000} = 31.622776601683793.
     $$

**Determine the Greatest Integer Not Exceeding $$ x $$:**

- The greatest integer not exceeding $$ x = 31.622776601683793 $$ is:
     $$
     \lfloor x \rfloor = 31.
     $$

### Final Answer:
The greatest integer not exceeding $$ x $$ is **31**.  
Correct option: **31**

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