How do you calculate # log_3 sqrt243#?

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Answer 1 · 2016-04-24T01:38:25

First, use the exponent rule #root(a)(n) = n^(1/a)#

#log_3(243^(1/2))#

Now, use the log rule #loga^n = nloga#

#1/2log_3(243)#

Now, use the log rule #log_an = logn/loga#

#1/2(log243/log3)#

Rewrite 243 in base 3.

#1/2((log3^5)/log3)#

#1/2((5log3)/log3)#

#1/2 xx 5#

#5/2#

Thus, completely simplified, #log_3(sqrt(243)) = 5/2 or 2.5#

Hopefully this helps!