CAT 2023 Slot 1 · Quant · Question 1

A logarithm equal to 4 means the base raised to the fourth power gives the argument.

So {\log}_{x}\left({x}^{2}+12 \right)=4 becomes {x}^{4}={x}^{2}+12.

Substituting u={x}^{2} turns this into the quadratic {u}^{2}-u-12=0.

Factorising, \left( u-4 \right)\left( u+3 \right)=0, so u=4 or u=-3.

A square cannot be negative, so u=4 and x=2 — taking the positive root since x is a positive real and a logarithm base.

The second condition, 3{\log}_{y}x=1, gives {\log}_{y}x=\frac{1}{3}.

That means {y}^{\frac{1}{3}}=x, so y={x}^{3}.

With x=2, y={2}^{3}=8.

Therefore x+y=2+8=10, which is the fourth option.

Shortcut. A logarithm base must be positive and not 1, so x=2 is the only candidate worth testing — and {2}^{4}=16={2}^{2}+12 checks instantly. Then {\log}_{y}x=\frac{1}{3} simply says y is the cube of x.