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CAT 2023 Slot 1 — QA Question 12

LogarithmsEasy

If x and y are positive real numbers such that logx (x2 + 12) = 4 and 3logy x = 1, then x + y equals?

Answer & solution

  • A

    11

  • B

    68

  • C

    20

  • 10

Solution

Easy

Convert each log equation to exponential form. The first becomes a quadratic in x2x^2 (solve, reject the impossible root); the second pins down yy directly. Then add.

1

Unpack log⁡x(x2+12)=4\log_x(x^2+12)=4: let a=x2a=x^2:

x2+12=x4 ⇒ a2−a−12=0⇒ (a−4)(a+3)=0 ⇒ a=4 or a=−3\begin{aligned} &x^2+12=x^4\ \Rightarrow\ a^2-a-12=0\\ &\Rightarrow\ (a-4)(a+3)=0\ \Rightarrow\ a=4\ \text{or}\ a=-3 \end{aligned}

Since a=x2≥0a=x^2\ge 0, reject −3-3, so x2=4x^2=4, and as a log base needs x>0x>0, x=2x=2.

2

Unpack 3log⁡yx=13\log_y x=1 with x=2x=2:

3log⁡y2=1 ⇒ log⁡y23=1⇒ y1=23 ⇒ y=8\begin{aligned} &3\log_y 2=1\ \Rightarrow\ \log_y 2^3=1\\ &\Rightarrow\ y^1=2^3\ \Rightarrow\ y=8 \end{aligned}
3

Add:

x+y=2+8=10x+y=2+8=10

x+y=10x+y=10 — option (d).

CAT 2023 Slot 1 QA Q12: If x and y are positive real numbers such that log x (x 2 + 12) = 4 and 3log y x = 1, then x + y equals? — Solution | TheCATExam