CAT 2024

Slot 1

QA

Question & Solution

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.