How do you evaluate $1/2 log_2 64 + 1/3 log_5 125$ ?

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Answer 1 · 2016-04-17T05:14:57

4

Use $log_aa=1 and log (a^m)=m log a$

The given expression is

$(1/2)log_2(2^6) + (1/3) log_5(5^3)=(1/2)6 log_2 2 + (1/3) 3 log_5 5=3+1=4$

How do you evaluate 1/2 log_2 64 + 1/3 log_5 1251/2 log_2 64 + 1/3 log_5 125?