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CAT 2018 Slot 1 — QA Question 23

LogarithmsEasy

If log1281 = p, then 3(4-p4+p) is equal to

Answer & solution

Correct answer: log 6 8

  • A

    log28

  • B

    log616

  • log68

  • D

    log416

Solution

Easy

Convert pp into log⁡123\log_{12}3, then simplify 4−p4+p\tfrac{4-p}{4+p} using log⁡1212=1\log_{12}12=1 to fold the numbers into logs of 44 and 3636. The expression collapses to a single log; multiplying by 33 gives a clean log.

1

Rewrite pp.

p=log⁡1281=log⁡1234=4log⁡123\begin{aligned} &p = \log_{12}81 = \log_{12}3^4 = 4\log_{12}3 \end{aligned}
2

Simplify the fraction. Replace each 44 using log⁡1212=1\log_{12}12=1 (so 4log⁡1212=44\log_{12}12=4):

4−p4+p=4−4log⁡1234+4log⁡123=1−log⁡1231+log⁡123⇒ =log⁡1212−log⁡123log⁡1212+log⁡123=log⁡124log⁡1236⇒ =log⁡124log⁡1236=log⁡364(change of base)\begin{aligned} &\frac{4-p}{4+p} = \frac{4-4\log_{12}3}{4+4\log_{12}3} = \frac{1-\log_{12}3}{1+\log_{12}3}\\ &\Rightarrow\ = \frac{\log_{12}12 - \log_{12}3}{\log_{12}12 + \log_{12}3} = \frac{\log_{12}4}{\log_{12}36}\\ &\Rightarrow\ = \frac{\log_{12}4}{\log_{12}36} = \log_{36}4 \quad\text{(change of base)} \end{aligned}
3

Reduce to base 66. Since 4=224=2^2 and 36=6236=6^2:

log⁡364=2log⁡22log⁡6=log⁡62⇒ 4−p4+p=log⁡62\begin{aligned} &\log_{36}4 = \frac{2\log 2}{2\log 6} = \log_6 2\\ &\Rightarrow\ \frac{4-p}{4+p} = \log_6 2 \end{aligned}
4

Multiply by 33.

3 ⁣(4−p4+p)=3log⁡62=log⁡623=log⁡68\begin{aligned} &3\!\left(\frac{4-p}{4+p}\right) = 3\log_6 2 = \log_6 2^3 = \log_6 8 \end{aligned}
3 ⁣(4−p4+p)=log⁡683\!\left(\dfrac{4-p}{4+p}\right) = \mathbf{\log_6 8}

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CAT 2018 Slot 1 QA Q23: If log 12 81 = p, then 3 4 - p 4 + p is equal to — Solution | TheCATExam