Question 22.

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.