calculus problem, help pleaseee :(

[Omegachi]

what is the inverse of:

y=LN(x+3)


and


y=2^10^x

 

[udonoogen]

<< what is the inverse of: y=LN(x+3) and y=2^10^x >>



to find the inverse of something, i think you have to switch the x and the y's around. then solve. thats what i remember from hs ... but as a poli sci major ... u dont have to take calculus. =)

first one i think is y = e^x - 3

could be wrong. im in humanities.

 

[Omegachi]

yea

the inverse is just switching the x and y, and yep y=e^x-3 is the answer but i don't know how you get that

 

[udonoogen]

y = ln (x+3)

inverse:
x = ln (y+3)
e^x = e^ (ln (y+3))
e^x = (y+3) ;; e and ln cancel each other
y = e^x - 3

 

[Omegachi]

thanks.....so e^ln cancels each other?

 

[udonoogen]

e^ln = 1
ln^e = 1

 

[Omegachi]

now for the y=2^10^x problem

 

[udonoogen]

mmm ... some identity. or log. i'd look it up but i think i should sleep. its 4 am here. what are you doing calculus for on a saturday night anyway. =P

 

[Omegachi]

haha, thanks. Its 2:30 here and i got nothing to do, so might as well do my homework.

 

[Shooters]

You just have to remember the property: ln (x^y) = y * ln(x)

So, to find the inverse of y = 2^10^x

x = 2^10^y
ln x = ln (2^10^y) = 10^y * ln 2
ln x / ln 2 = 10^y
ln (ln x / ln 2) = ln (10^y) = y * ln 10
y = (ln (ln x / ln 2)) / ln 10

There's probably a way to simplify that last expression, but I can't think of it at the moment other than using the property: ln (y/x) = ln y - ln x

 

[Omegachi]

wow dude..you are a genius!

 

[rival]

man wtf, i got "74"

 

[Shooters]

<< wow dude..you are a genius! >>


No.....just an engineering student who has had to take that class before.