Given the equation: $4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$. We need to find the greatest integer less than or equal to \( x \).
First, simplify the logarithmic terms using the change of base formula:
- \(\log_{100} x = \log_{10^2} x = \frac{1}{2} \log_{10} x\)
- \(\log_{1000} x = \log_{10^3} x = \frac{1}{3} \log_{10} x\)
Substitute these into the original equation:
\(4 \log_{10} x + 4 \left(\frac{1}{2} \log_{10} x\right) + 8 \left(\frac{1}{3} \log_{10} x\right) = 13\)
Simplify the equation by performing the multiplications:
- \(4 \log_{10} x + 2 \log_{10} x + \frac{8}{3} \log_{10} x = 13\)
Combine the logarithmic terms:
\(\left(4 + 2 + \frac{8}{3}\right) \log_{10} x = 13\)
Calculate the sum of the coefficients:
\(\frac{12}{3} + \frac{6}{3} + \frac{8}{3} = \frac{26}{3}\)
The equation simplifies to:
\(\frac{26}{3} \log_{10} x = 13\)
To isolate \(\log_{10} x\), multiply both sides by \(\frac{3}{26}\):
\(\log_{10} x = \frac{13 \cdot 3}{26} = \frac{39}{26} = \frac{3}{2}\)
Convert the logarithmic equation to its exponential form to solve for \( x \):
\(x = 10^{\frac{3}{2}}\)
Rewrite \( x \) in radical form:
\(x = \sqrt{10^3} = \sqrt{1000} = 10 \sqrt{10}\)
Using the approximation \(\sqrt{10} \approx 3.162\), calculate the approximate value of \( x \):
\(x \approx 10 \times 3.162 = 31.62\)
The greatest integer not exceeding \( x \) is:
\(\boxed{31}\)
The computed value of \(x \approx 31.62\) falls within the range [31, 32). Therefore, the greatest integer less than or equal to \( x \) is 31.