How do you evaluate #log_5 (5^-1)#?

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Answer 1 · 2016-10-29T15:08:04

Use the rule #loga^n = nloga#.

#= -1log_5(5)#

Use the change of base rule #log_a(n) = logn/loga#.

#=-1log5/log5#

#=-1#

Hopefully this helps!

Answer 2 · 2016-10-29T15:24:58

#log_5(5^(-1))#

#=-log_5(5)#

#=-1#

This is because:

#log_a(m^n)=x#

Which means that:

#a^x=m^n#

Then:

#a^(x/n)=m#

As a result:

#log_a(m)=x/n#

And this implies that:

#x=n*log_a(m)#

Now, considering the fact that #x=log_a(m^n)# as stated at the beginning of this proof, we can say that:

#n*log_a(m)=log_a(m^n)#

And this is why #log_5(5^-1)# evaluated is equal to #-1#.