# Thread: ln(1/2) & ln(e^k^t)

1. ## ln(1/2) & ln(e^k^t)

$\displaystyle \frac{1}{2}=e^k^t$

When ln() is taken to both sides of the equation it is supposed to simplify to:

$\displaystyle -ln(2)=kt$

I'm wondering by what rules was each side simplified. I know that ln(e) is 1 but in this case ln(e^k^t) is the quantity k*t? The fact that ln(1/2) simplifies somehow to -ln(2) just leaves me bewildered at this point.

Thanks in advance for any posts on this...

2. ## Re: ln(1/2) & ln(e^k^t)

Originally Posted by sepoto
$\displaystyle \frac{1}{2}=e^k^t$

When ln() is taken to both sides of the equation it is supposed to simplify to:

$\displaystyle -ln(2)=kt$

I'm wondering by what rules was each side simplified. I know that ln(e) is 1 but in this case ln(e^k^t) is the quantity k*t? The fact that ln(1/2) simplifies somehow to -ln(2) just leaves me bewildered at this point.

Thanks in advance for any posts on this...
You know that
$\displaystyle ln(a^n) = n \cdot ln(a)$

So write $\displaystyle \frac{1}{2} = 2^{-1}$. So what is $\displaystyle ln(2^{-1})$?

-Dan

Edit: As to your other question, the RHS is $\displaystyle e^{kt}$, not e^{k^t}. So $\displaystyle ln(e^{kt}) = kt \cdot ln(e) = kt$.

3. ## Re: ln(1/2) & ln(e^k^t)

Another way to look at ln(1/2): ln(A/B)= ln(A)- ln(B) so ln(1/2)= ln(1)- ln(2)= 0- ln(2)= -ln(2).