Logarithmic Equation Floor Value

Change of base formulas: $\log_{100} x = \frac{\log_{10} x}{2}, \quad \log_{1000} x = \frac{\log_{10} x}{3}$

Substitute into the equation: $4\log_{10} x + 4\left(\frac{\log_{10} x}{2}\right) + 8\left(\frac{\log_{10} x}{3}\right) = 13$ $\left(4 + 2 + \frac{8}{3}\right) \log_{10} x = 13$ $\frac{26}{3} \log_{10} x = 13 \implies \log_{10} x = \frac{3}{2}$ $x = 10^{3/2} = 10\sqrt{10} \approx 31.62$

Greatest integer not exceeding $x$ is $\lfloor 31.62 \rfloor = 31$.