TheCATExam
Sign in

CAT 2020 Slot 1 — QA Question 23

LogarithmsEasy

If y is a negative number such that 2y2log35=5log23, then y equals

Answer & solution

  • A

    log₂(1/5)

  • B

    –log₂ (1/3)

  • C

    –log₂ (1/5)

  • log₂ (1/3)

Solution

Easy

The unknown yy sits in an exponent, so take logarithms of both sides. Choosing base 33 makes the log⁡35\log_3 5 terms cancel cleanly, leaving a pure equation in y2y^2. Then use the sign condition (y<0y<0) to fix the root.

1

Take log⁡3\log_3 of both sides. Apply log⁡3\log_3 to 2 y2log⁡35=5 log⁡232^{\,y^2\log_3 5}=5^{\,\log_2 3} and bring exponents down.

2 y2log⁡35=5 log⁡23⇒ y2 (log⁡35)(log⁡32)=(log⁡23)(log⁡35)(took log⁡3 both sides)\begin{aligned} &2^{\,y^2\log_3 5}=5^{\,\log_2 3}\\ &\Rightarrow\ y^2\,(\log_3 5)(\log_3 2)=(\log_2 3)(\log_3 5) \quad\text{(took $\log_3$ both sides)} \end{aligned}
2

Cancel log⁡35\log_3 5. It is nonzero, so divide it out from both sides.

y2 log⁡32=log⁡23(cancel log⁡35)⇒ y2=log⁡23log⁡32=(log⁡23)2(since log⁡32=1log⁡23)\begin{aligned} &y^2\,\log_3 2=\log_2 3 \quad\text{(cancel $\log_3 5$)}\\ &\Rightarrow\ y^2=\frac{\log_2 3}{\log_3 2}=(\log_2 3)^2 \quad\text{(since $\log_3 2=\tfrac{1}{\log_2 3}$)} \end{aligned}
3

Apply the sign condition. Since yy is negative, take the negative square root.

y=−log⁡23(because y<0)⇒ y=log⁡2 ⁣(13)(since −log⁡23=log⁡23−1)\begin{aligned} &y=-\log_2 3 \quad\text{(because $y<0$)}\\ &\Rightarrow\ y=\log_2\!\Big(\tfrac{1}{3}\Big) \quad\text{(since $-\log_2 3=\log_2 3^{-1}$)} \end{aligned}
y=log⁡2 ⁣(13)y=\log_2\!\left(\tfrac{1}{3}\right)
CAT 2020 Slot 1 QA Q23: If y is a negative number such that 2 y 2 log 3 5 = 5 log 2 3 , then y equals — Solution | TheCATExam