Mathematics
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OpenStudy (osanseviero):
Inverse logarithmic function
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OpenStudy (osanseviero):
So they want the inverse of \[F(x)=\ln \sqrt{x}\]
So the inverse is
\[x=\ln \sqrt{y}\]
hartnn (hartnn):
yep,
now isolate y
hartnn (hartnn):
\(\Large a=\ln c \implies c = e^a \)
OpenStudy (osanseviero):
\[y=\left( e ^{x} \right)^{2}\] the 2 can go down, right?
OpenStudy (osanseviero):
y=2e^x
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hartnn (hartnn):
not actually
hartnn (hartnn):
\(\Large (e^x)^2 = e^{2x}\)
OpenStudy (osanseviero):
oops, another silly mistake
OpenStudy (osanseviero):
Can someone help me demostrate that (FoFinverse)(x)=(FinverseoF)(x)=x ?
hartnn (hartnn):
demonstrate ?
you have a function for verifying that ?
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OpenStudy (osanseviero):
The one we just did
OpenStudy (osanseviero):
\[\ln \sqrt{e ^{2x}} = e ^{2\ln \sqrt{x}}=x\]
hartnn (hartnn):
so, f inverse = e^2x
just plug in x =ln \sqrt x
oh, you did it :)
OpenStudy (osanseviero):
But how and why? (I think that was a good statement)
OpenStudy (osanseviero):
And how would that be x?
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hartnn (hartnn):
ln x^m = m ln x
hartnn (hartnn):
ln e =1
hartnn (hartnn):
\(\sqrt {(e^x)^2} = |e^x| \\ \ln |e^x| = x \ln e = x\)
hartnn (hartnn):
ok ?
OpenStudy (osanseviero):
so the left one is kinda obvious, so it is x
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OpenStudy (osanseviero):
Oh...and the other one is with
\[a ^{logaN}=\]
OpenStudy (osanseviero):
N
hartnn (hartnn):
\(\huge a^{\log_aN}=N\)
hartnn (hartnn):
yes
OpenStudy (osanseviero):
:) Now I understand, thanks for fifth time, just one more question :P
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hartnn (hartnn):
welcome ^_^
OpenStudy (osanseviero):
Which would be the domain and the range of the first, normal function?
hartnn (hartnn):
domain of ln sqrt x ?
hartnn (hartnn):
since its sqrt x, x>= 0
since its ln, sqrt x >0
in all, x>0
OpenStudy (osanseviero):
so the range of the inverse is x>0
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OpenStudy (osanseviero):
Wish me good luck tomorrow
hartnn (hartnn):
yes!
best of luck! hope you get full marks :)
OpenStudy (osanseviero):
I hope so, this will be a hard examn...all types of functions, trigonometric identities, operations with functions
hartnn (hartnn):
if you have practices enough, nothing is hard :)
OpenStudy (osanseviero):
Thanks for the help all night!
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hartnn (hartnn):
you're welcome ^_^