anonymous
  • anonymous
Given f(x)=log11x, find f–1(x) and f–1(2).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[given f(x)=\log_{11} x, find f ^{-1}(x) and f ^{-1}(2)\]
anonymous
  • anonymous
Can anyone help?
SolomonZelman
  • SolomonZelman
u can use ~ for space

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SolomonZelman
  • SolomonZelman
An inverse of logarithmic is going to be an exponential.
anonymous
  • anonymous
Inverse?
SolomonZelman
  • SolomonZelman
\({\rm f}^{-1}(x)\) is a notation for an inverse function.
anonymous
  • anonymous
Oh okay.. I suck at these to be honest
SolomonZelman
  • SolomonZelman
Your function is: \(\large\color{black}{ \displaystyle f(x)=\log_{11}(x) }\) 1) replace f(x) with y. 2) switch x and y (write y instead of x, and write y instead of x) 3) {solve for/ isolate the} y.
anonymous
  • anonymous
\[x=\log_{11} y\]
anonymous
  • anonymous
like that?
SolomonZelman
  • SolomonZelman
for the third step, you will need to know that \(\Large\color{black}{ \displaystyle \color{red}{\rm a}=\log_{\color{green}{\rm b}}(\color{blue}{\rm c}) ~~~~\Longrightarrow~~~~\color{green}{\rm b}^\color{red}{\rm a}=\color{blue }{\rm c} }\).
SolomonZelman
  • SolomonZelman
yes, for the first two steps, that is correct
SolomonZelman
  • SolomonZelman
now, apply the third step to that (to \(x=\log_{11}y\) )
anonymous
  • anonymous
\[11^{x}\]
anonymous
  • anonymous
=y
anonymous
  • anonymous
\[11^{x}=y\] Like that? @SolomonZelman
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
then replcae y by \(f^{-1}(x)\)
SolomonZelman
  • SolomonZelman
|dw:1435782711063:dw|
SolomonZelman
  • SolomonZelman
then for f\(^{-1}\)(2) plug in 2 for x, into the last expression that I drew
anonymous
  • anonymous
where did you get the 2 from?
SolomonZelman
  • SolomonZelman
you need \(f^{-1}(2)\). that is your second part.
anonymous
  • anonymous
ohhh okay I got it now
anonymous
  • anonymous
Can you help me with another one? @SolomonZelman
SolomonZelman
  • SolomonZelman
ok, but I am helping another person right now... well I am attempting to help that person.
SolomonZelman
  • SolomonZelman
will see.
anonymous
  • anonymous
Okay.. I'll just open a new question.

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