How do you calculate #Log_9 49#?

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Answer 1 · 2016-06-01T14:26:37

#y = log_3 7#

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#