cat 2023 Question 1

Question

If x and y are positive real numbers such that $\log _x\left(x^2+12\right)=4$ and $3 \log _y x=1$ , then x+y equals

Accepted Answer

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(x^2 - 4) + 3(x^2 - 4) = 0 $

$ (x^2 - 4)(x^2 + 3) = 0 \implies$ since $x$ is a positive real number, $x = 2$

Now, given $ 3 \log_{y} x = 1 $

$ \log_{y} x = \frac{1}{3} $

$ x = y^{\frac{1}{3}} $

$ y = x^3 $

$ y = 8 $

$ x + y = 2 + 8 = 10 $

Question 1.