CAT 2018 Slot 2QA Question 22

LogarithmsEasy

The smallest integer n for which 4n > 1719 holds, is closest to

Answer & solution

  • A

    33

  • B

    37

  • 39

  • D

    35

Solution

Easy

Take logs to base 44. Since 1717 is just above 16=4216=4^2, log417\log_4 17 is just over 22, so nn must just exceed 2×19=382\times 19=38 — the smallest integer is 3939.

1

Take log4\log_4 of both sides.

4n>1719 n>19log417\begin{aligned} &4^{n}>17^{19}\\ &\Rightarrow\ n>19\,\log_4 17 \end{aligned}
2

Estimate log417\log_4 17. Since 42=16<174^2=16<17, log417\log_4 17 is slightly more than 22 (about 2.0432.043).

log4172.043 19log41738.8\begin{aligned} &\log_4 17\approx 2.043\\ &\Rightarrow\ 19\,\log_4 17\approx 38.8 \end{aligned}
3

Smallest integer above the bound.

n>38.8 nmin=39\begin{aligned} &n>38.8\\ &\Rightarrow\ n_{\min}=39 \end{aligned}
n=39n=39

Check: 439=2784^{39}=2^{78} and 1719<(24.1)19=277.917^{19}<(2^{4.1})^{19}=2^{77.9}, while 438=276<17194^{38}=2^{76}<17^{19}. So 3939 is indeed the first integer to clear the bound.

CAT 2018 Slot 2 QA Q22: The smallest integer n for which 4 n > 17 19 holds, is closest to — Solution | TheCATExam