How do you solve #log_11(sqrt(11^7))=n#?

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Answer 1 · 2015-11-30T22:50:58

#n=7/2#

The first step to take is rewriting #sqrt(11^7)# as #(11^7)^(1/2)#, which equals #11^(7/2)#.

So, we now have:

#log_11 11^(7/2)=n#

Now, we can use the following rule: #log_a b^c=clog_a b#.

We get:

#7/2log_11 11=n#

Since #log_a a=1#, we know that #log_11 11=1#.

Therefore, #n=7/2#.