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CAT 2019 Slot 1 — QA Question 31

LogarithmsEasy

Let x and y be positive real numbers such that log5(x + y) + log5(x - y) = 3, and log2y - log2x = 1 - log23. Then xy equals

Answer & solution

Correct answer: 150

  • A

    250

  • 150

  • C

    100

  • D

    25

Solution

Easy

Combine the base-5 logs into a single product and clear the log to get x2−y2x^2-y^2. Combine the base-2 logs to get a linear relation between xx and yy. Solve the two equations together for x,yx,y, then multiply.

1

First equation (base 5). Use log⁡a+log⁡b=log⁡(ab)\log a+\log b=\log(ab).

log⁡5(x+y)+log⁡5(x−y)=3⇒ log⁡5[(x+y)(x−y)]=3⇒ x2−y2=53=125...(1)\begin{aligned} &\log_5(x+y)+\log_5(x-y)=3\\ &\Rightarrow\ \log_5\big[(x+y)(x-y)\big]=3\\ &\Rightarrow\ x^2-y^2=5^3=125\quad\text{...(1)} \end{aligned}
2

Second equation (base 2). Note 1−log⁡23=log⁡2231-\log_2 3=\log_2\frac{2}{3}.

log⁡2y−log⁡2x=1−log⁡23⇒ log⁡2yx=log⁡223⇒ yx=23 ⇒ 3y=2x...(2)\begin{aligned} &\log_2 y-\log_2 x=1-\log_2 3\\ &\Rightarrow\ \log_2\frac{y}{x}=\log_2\frac{2}{3}\\ &\Rightarrow\ \frac{y}{x}=\frac{2}{3}\ \Rightarrow\ 3y=2x\quad\text{...(2)} \end{aligned}
3

Solve (1) and (2). From (2), x=3y2x=\tfrac{3y}{2}; substitute into (1).

(3y2)2−y2=125[from (1),(2)]⇒ 9y24−y2=125 ⇒ 5y24=125⇒ y2=100 ⇒ y=10,x=3(10)2=15\begin{aligned} &\left(\tfrac{3y}{2}\right)^2-y^2=125\quad\text{[from (1),(2)]}\\ &\Rightarrow\ \tfrac{9y^2}{4}-y^2=125\ \Rightarrow\ \tfrac{5y^2}{4}=125\\ &\Rightarrow\ y^2=100\ \Rightarrow\ y=10,\quad x=\tfrac{3(10)}{2}=15 \end{aligned}
4

Compute the product.

xy=15×10=150\begin{aligned} &xy=15\times 10=150 \end{aligned}
xy=150xy=150

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CAT 2019 Slot 1 QA Q31: Let x and y be positive real numbers such that log 5 (x + y) + log 5 (x - y) = 3, and log 2 y - log 2 x = 1 - — Solution | TheCATExam