# Thread: Proving ln( 5 root e) = 1/5

1. ## Proving ln( 5 root e) = 1/5

I need help with this proof...

Prove that $\displaystyle ln 5 sqrt{e} = 1/5$

It is actually like ln e^(1/5) but I dont know how to make the 5 small...

2. Originally Posted by circuscircus
I need help with this proof...

Prove that $\displaystyle ln 5 sqrt{e} = 1/5$

It is actually like ln e^(1/5) but I dont know how to make the 5 small...
You mean ln[fifth root of e].
"...small 5 sqrt...." ----------that is fifth root.

So, prove ln[e^(1/5)] = 1/5

We will use:
---Log[a^b] = b*Log[a]
---Log(base a)[a] = 1

ln[e^(1/5)]
= (1/5)ln[e]
= (1/5)*Log(base e)[e]
= (1/5)*1
= 1/5

3. Where does this come from?

Log[a^b] = b*Log[a]

4. Originally Posted by circuscircus
Where does this come from?

Log[a^b] = b*Log[a]
Recall that
$\displaystyle log(n \cdot a) = log(n) + log(a)$

What does $\displaystyle a^b$ mean? It means $\displaystyle a^b = a\cdot a \cdot ~ ... ~ \cdot a$ where we are doing this "b" times.

So
$\displaystyle log(a^b) = log(a) + log(a) + ~ ... ~ + log(a) = b \cdot log(a)$

Note that it does not matter what base the logarithm is to:
$\displaystyle log_{10}(a^b) = b \cdot log_{10}(a)$
and
$\displaystyle log_2(a^b) = b \cdot log_2(a)$

-Dan