How do you simplify #4^(log_4 64) + 10^(log 100)#?

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Answer 1 · 2016-11-15T14:44:41

It can be evaluated to #164#.

Recall that #log_a(n) = logn/loga#.

#=>4^(log64/log4) + 10^(log100/log10)#

#=>4^(log4^3/log4^1) + 10^(log10^2/log10^1)#

#=>4^((3log4)/(1log4)) + 10^((2log10)/(1log10))#

#=> 4^3 + 10^2#

#=> 64 + 100#

#=> 164#

Hopefully this helps!