How do I simplify e^[-4 ln (tanϴ)]?

Question

anonymous · Answer

Hi @baru. Any idea? :) I have a feeling that we should be focusing on the -4 ln (tanϴ) part but I'm not sure...

baru · Answer

consider z=ln( tanϴ)

then e^z=tan(ϴ)

anonymous · Answer

I'm still a bit confused, I'm really not sure what to do!

baru · Answer

$$e^{-4\ln(\tan \theta)}=e^{\ln(\tan^{-4} \theta)}$$

baru · Answer

$$Let ~ \ln(\tan^{-4}\theta)=z\ e^z=(\tan^{-4}\theta)$$

anonymous · Answer

Okay I understand it so far :)

baru · Answer

$$e^{\ln(\tan^{-4} \theta)}$$

substitute 'z' in the  above equation wherever possible

baru · Answer

so we just get $$e^z$$

anonymous · Answer

I'm just confused because can't we just do ln[tan(theta)]/e^4?

baru · Answer

no you cant$$a^{-bc}\neq \frac{c}{a^b}$$

what i've typed above is what you tried to do

anonymous · Answer

Oh I see.

anonymous · Answer

Can I read my notes again and come back later? Thanks for all your help by the way. I just want to make sure I really understand this content!

baru · Answer

sorry if i confused you...

the main thing to remember is this rule
$$e^{\ln(a)}=a$$

anonymous · Answer

Thanks, I'll look into it. And don't worry, perhaps I just need to make sure I really understand this material! :)

baru · Answer

sure :)

kenshin · Answer

Baru I just typed both e^(-4lntan(x)) and (tan(x))^-4 into my graphing calculator; the graphs are slightly different, it appears whilst certain parts are the same, the exponent version has some parts missing, is this because there are domain restrictions for x inside ln?

baru · Answer

hmm curious...
i entered the starting expression into wolframalpha, the answer seems correct

kenshin · Answer

tan(x) cannot be between (-infinity, 0) whilst inside the LN?

kenshin · Answer

Wolfram seems to have same graph for both for me too.

baru · Answer

yea ur right, tan cannot take negative values here, 
in any case, this is an exercise to get one comfortable with laws of exponents and logarithms.. so OP need not worry :)

baru · Answer

tan will take negative values in the 2nd and 4th quadrants, so i guess parts of the graph will  go missing periodically

baru · Answer

then again, a negative number raised to an even power is positive...
so perhaps you typed it slightly differently in wolfram and in your calculator

baru · Answer

the question exactly the way it appears is undefined in some parts

however in
$$e^{\ln(\tan^{-4} \theta)}$$
tan is raised to an even power

kenshin · Answer

ok that's right, so tan(x) cannot be zero only, negative values can be valid. So x cannot be k(pi) where k = integers and 0, all other values are ok.

baru · Answer

i guess... but i'm no expert :p