How do you solve #log_2 n = 1/3 log_2 27 + log_2 36#?

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Answer 1 · 2016-06-01T21:02:51

#log_2(n) = 1/3log_2(27) + log_2(36)#

Use the rule #alogn = logn^a# to get rid of #log_2(27)#'s coefficient.

#log_2(n) = log_2(27^(1/3)) + log_2(36)#

Now, know that #a^(1/n) = root(n)(a)#

#log_2(n) = log_2(3) + log_2(36)#

Use the rule #log_a(n) + log_a(m) = log_a(n xx m)# to make the right side into one logarithm.

#log_2(n) = log_2(3 xx 36)#

#log_2(n) = log_2(108)#

#n = 108#

Hopefully this helps!