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

2 Answers
Oct 29, 2016

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!

Oct 29, 2016

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.