How do you calculate #Log_9 49#?

1 Answer
Jun 1, 2016

Answer:

#y = log_3 7#

Explanation:

We know that #y = log_9 49 equiv 9^y = 49#

Supposing you have for computational purposes the function

#log_b#, with #b# the basis, you can proceed as
#log_b 9^y = log_b 49->y log_b 9 = log_b 49#

solving for #y#

#y = (log_b 49)/(log_b 9) = (log_b 7^2)/(log_b 3^2) = (log_b 7)/(log_b 3)#.

If you choose #b = 3# as basis then the result will read

#y = log_3 7#