CAT 2024 Slot 1 · Quant · Question 22

Put every logarithm on base 10, using {\log}_{{10}^{k}}x=\frac{{\log}_{10}x}{k}.

Write L={\log}_{10}x. Then {\log}_{100}x=\frac{L}{2} and {\log}_{1000}x=\frac{L}{3}.

The equation becomes 4L+4\cdot \frac{L}{2}+8\cdot \frac{L}{3}=13.

That is 4L+2L+\frac{8L}{3}=13.

Combining, 6L+\frac{8L}{3}=\frac{18L+8L}{3}=\frac{26L}{3}.

So \frac{26L}{3}=13, giving L=\frac{39}{26}=\frac{3}{2}.

Then x={10}^{\frac{3}{2}}=10\sqrt{10}\approx 31.62.

The greatest integer not exceeding it is 31.

Shortcut. A base that is a power of 10 just divides the logarithm, so the three terms are 4L, 2L and \frac{8L}{3} on sight. Note also that 10\sqrt{10} sits between 31 and 32 because \sqrt{10}\approx 3.162, so no decimal work is needed.