CAT 2023

Slot 1

QA

Question & Solution

Given, $\log_{x}(x^2 + 12) = 4$

=> $x^2+12=x^4$

=> $x^4-x^2-12=0$

=> $x^4-4x^2+3x^2-12=0$

=> $x^2\left(x^2-4\right)+3\left(x^2-4\right)=0$

=> $\left(x^2-4\right)\left(x^2+3\right)=0$ => since, x is a positive real number (given) => x = 2.

Now, Given $3 \log_{y} x = 1$

=> $\log_yx=\frac{1}{3}$

=> $x=y^{\frac{1}{3}}$

=> $y=x^3$ => y = 8.

=> x + y = 2 + 8 = 10.