If #log_a 36=2.2265# and #log_a 4=.8614#, how do you evaluate #log_a 9#?

Question

1816 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-11T17:51:58

Using the property of #log# which states that #log_n(a/b)=log_na-log_nb#

Following the #log# property, we can use your pieces of information as follows:

#log9=log(36/4)#

Thus,

#log_a36-log_a4=log_a9#

#2.2265-0.8614=log_a9#

#log_a9=1.3651#

Following #log# definition, we have that #log_ab=c <=> a^c=b#

Thus,

#a^(1.3651)=9#

To solve this, isolating #a# we can follow one property of exponentials, which states #(a^n)^m=a^(n*m)# by elevating both sides to #(1/1.3651)#:

#(a^(cancel(1.3651)))^(1/cancel(1.3651))=9^(1/1.3651)#

#a=9^(1/1.3651)#

#a~=0.7325#