CAT 2024 Slot 3 · Quant · Question 22

The same quantity {10}^{x} appears twice, so substitute y={10}^{x}, where y>0.

The equation becomes y+\frac{4}{y}=\frac{17}{2}.

Multiplying through by 2y to clear the fractions, 2{y}^{2}+8=17y.

Rearranging, 2{y}^{2}-17y+8=0.

The discriminant is {17}^{2}-4\times 2\times 8=289-64=225, whose square root is 15.

So y=\frac{17\pm 15}{4}, giving y=8 or y=\frac{1}{2}.

Both are positive, so both are legitimate values of {10}^{x}.

From {10}^{x}=8, x={\log}_{10}8; from {10}^{x}=\frac{1}{2}, x={\log}_{10}\frac{1}{2}.

Adding, the sum is {\log}_{10}8+{\log}_{10}\frac{1}{2}={\log}_{10}\left( 8\times \frac{1}{2}\right)={\log}_{10}4.

Since 4={2}^{2}, this is 2{\log}_{10}2 — the first option.

Shortcut. The two roots multiply to \frac{8}{2}=4 by Vieta, and the sum of the logarithms is the logarithm of the product. So the answer is {\log}_{10}4=2{\log}_{10}2 without ever naming the individual roots.