How do you combine #\log _ { 9} 36- \log _ { 3} 2#?

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Answer 1 · 2017-07-28T07:38:37

See explanation.

In the given expression there are 2 logarithms with different bases. To combine such expression we have to use the base change formula:

#log_c a=log_ba/log_bc# ##

Here we can change base 9 logarithm to base 3:

#log_9 36=log_3 36/log_3 9=log_3 36/2#

So the expression becomes:

#log_3 36/2-log_3 2#

#=1/2log_3 36-log_3 2" "larr# use the power law

#=log_3 36^(1/2)-log_3 2" "larr (x^(1/2) = sqrtx rarr sqrt36 =6)#

#=log_3 6-log_3 2=#

#=log_3 (6/2)=log_3 3=1#