How do you solve #log_3 42-log_3n=log_3 7#?
1 Answer
Apr 30, 2018
Explanation:
We should use
#log_a n - log_a m = log_a(n/m)# .
Therefore:
#log_3(42/n) = log_3 7#
#42/n = 7#
#7n =42#
#n = 6#
Hopefully this helps!
