How do you evaluate ${log}_{5} (5^{- 1})$ $log_5 (5^-1)$ ?

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Answer 1 · 2016-10-29T15:08:04

Use the rule ${log a}^{n} = n log a$ $loga^n = nloga$ .

$= - 1 {log}_{5} (5)$ $= -1log_5(5)$

Use the change of base rule ${log}_{a} (n) = \frac{log n}{log a}$ $log_a(n) = logn/loga$ .

$= - 1 \frac{log 5}{log 5}$ $=-1log5/log5$

$= - 1$ $=-1$

Hopefully this helps!

Answer 2 · 2016-10-29T15:24:58

${log}_{5} (5^{- 1})$ $log_5(5^(-1))$

$= - {log}_{5} (5)$ $=-log_5(5)$

$= - 1$ $=-1$

This is because:

${log}_{a} (m^{n}) = x$ $log_a(m^n)=x$

Which means that:

$a^{x} = m^{n}$ $a^x=m^n$

Then:

$a^{\frac{x}{n}} = m$ $a^(x/n)=m$

As a result:

${log}_{a} (m) = \frac{x}{n}$ $log_a(m)=x/n$

And this implies that:

$x = n \cdot {log}_{a} (m)$ $x=n*log_a(m)$

Now, considering the fact that $x = {log}_{a} (m^{n})$ $x=log_a(m^n)$ as stated at the beginning of this proof, we can say that:

$n \cdot {log}_{a} (m) = {log}_{a} (m^{n})$ $n*log_a(m)=log_a(m^n)$

And this is why ${log}_{5} (5^{- 1})$ $log_5(5^-1)$ evaluated is equal to $- 1$ $-1$ .

How do you evaluate log5(5−1)log_5 (5^-1)?