Domain of the inverse help! The function is f(x)=ln(e^y - 3).

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Domain of the inverse help! The function is f(x)=ln(e^y - 3).

Mathematics
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|dw:1443324096677:dw|
Ok so this is a one-to-one function
|dw:1443324955839:dw| this is what i got.. am i right?

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\[f(x) = \ln(e^x-3) \]\[y=\ln(e^x-3)\]\[x=\ln(e^y-3)\]\[\large e^x =e^{\ln_e(e^y-3)}\]\[e^x =e^y-3\]\[e^y=e^x+3\]\[\ln (e^y) = \ln(e^x+3)\]\[y=\ln(e^x+3)\] This is what I'm getting.
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oh yeah i have to use ln not log :3 thanks!
Uhh... not entirely.
\[\color{red}{\log_e (x) \equiv \ln(x)}\]
??
Either you write your inverse as a log based function, which entails it has a base of e, or you take the natural log function which works off base e as well, theyre equivalent
yeah, so if I write it in terms of log, it will be like: |dw:1443325371555:dw|
Yes
yay, thank you so much!
For practice, you should try this problem: \(\large f(x) =\dfrac{e^{2x}-1}{2-e^2x}\)
\[\large f(x) = \frac{e^{2x}-1}{2-e^{2x}}\]
ok, I will also find the inverse of that function?
Yeah. It'll help you practice how to find inverse of logarithmic functions
ok..|dw:1443325598398:dw||dw:1443325667079:dw| Can I simplify it more?
wait i think i did something wrong..
no i think i'm right :P |dw:1443326011460:dw|

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