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

Using the logarithmic property that $$\log_{a^p}b=\frac{1}{p}\log_ab$$

$$4 \log_{10} x + 4 \log_{100} x + 8 \log_{1000} x = 13$$
Can be written as 
$$4\log_{10}x+2\log_{10}x+\frac{8}{3}\log_{10}x=13$$
$$\frac{26}{3}\log_{10}x=13$$
$$\log_{10}x=1.5$$
$$x=10^{1.5}$$
$$x=\sqrt{1000}$$
$$\left[\sqrt{1000}\right]=31$$
Where [.] is Greatest Integer Function since that is what is asked in the question, 
31 is the greatest integer that does not exceed x.