It is more consistent to use log and compare. 3^200 = 2^log_2(3^200) = 2^(200*log_2(3)) And you need a calc for this method but in this case we can know that log_2(3) is bigger than 1.5 and 200*1.5 = 300 2^1.5 = 2 *sqrt(2) < 3 meaning 200 * log2(3) > 300
Thanks for watching, please like and don't forget to subscribe if you liked this video.
It is more consistent to use log and compare.
3^200 = 2^log_2(3^200) = 2^(200*log_2(3))
And you need a calc for this method but in this case we can know that log_2(3) is bigger than 1.5 and 200*1.5 = 300
2^1.5 = 2 *sqrt(2) < 3 meaning
200 * log2(3) > 300
👍