- Isolate ...

Question

- Isolate ...

asylum15 · Answer

$$e^ - {\frac{ 20 }{ \mu}} = \ln 0.9$$

How do I isolate mu? 

Mu = ?

asylum15 · Answer

That is the exponential of NEGATIVE 20 over mu.

ganeshie8 · Answer

take ln both sides

ganeshie8 · Answer

$\large   e^{-\frac{20}{\mu}} =  \ln 0.9$

take ln both sides 

$\large   \ln   (e^{-\frac{20}{\mu}}) =  \ln(\ln 0.9)$

asylum15 · Answer

Left side Ln + e cancel?

ganeshie8 · Answer

yup, and right side is just a constant

asylum15 · Answer

So how will the final line look? Just to clarify?

ganeshie8 · Answer

$\large   \ln   (e^{-\frac{20}{\mu}}) =  \ln(\ln 0.9)$


$\large   -\frac{20}{\mu}\ln   (e) =  \ln(\ln 0.9)$

$\large   -\frac{20}{\mu} =  \ln(\ln 0.9)$

$\large   -\frac{20}{\ln(\ln 0.9)} =  \mu$

asylum15 · Answer

My lecturer explained it horribly, but she went from the first line i had above to:

$$\mu = \frac{ -20 }{ \ln0.9 }$$

asylum15 · Answer

Is that the same as what you've done?

asylum15 · Answer

Since Ln (Ln) is a non process?

ganeshie8 · Answer

ln(ln(ln(lnx)))) is okay to process

ganeshie8 · Answer

however, let me ask u a q :-

is ur original q :
$e^ - {\frac{ 20 }{ \mu}} = \ln 0.9 $

?


or


$e^ - {\frac{ 20 }{ \mu}} = 0.9$

?

asylum15 · Answer

e−20μ=ln0.9

ganeshie8 · Answer

actually you're right, we cannot process ln(ln(0.9))
cuz ln(0.9) gives u negative number

ganeshie8 · Answer

looks u need to mess wid imaginaires http://www.wolframalpha.com/input/?i=e%5E%7B-%5Cfrac%7B20%7D%7B%5Cmu%7D%7D+%3D++%5Cln+0.9