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I am confused as to how the inverse of a certain function is found.

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2

$begingroup$

The problem within my textbook is $f(x)=$$frac3+lnx3-lnx$ I checked the answer and it is $y=$$e^frac3x-3x+1$
I've tried many times to simplify the equation after switching the variables but I don't know how to seperate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.

algebra-precalculus

share|cite|improve this question

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Could you type out what you have tried so far so we can try to spot your error?
  $endgroup$
  – TM Gallagher
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^ln y = e^....something to do with (x)....$.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac 3+ln y3-ln yimplies ln y = frac 3x -3x+1$?
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The problem within my textbook is $f(x)=$$frac3+lnx3-lnx$ I checked the answer and it is $y=$$e^frac3x-3x+1$
I've tried many times to simplify the equation after switching the variables but I don't know how to seperate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.

algebra-precalculus

share|cite|improve this question

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Could you type out what you have tried so far so we can try to spot your error?
  $endgroup$
  – TM Gallagher
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you have $ln y = ....something to do with(x)...$ then just raise $e$ to both sides of the equation. $y = e^ln y = e^....something to do with (x)....$.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  .... but you need to get $ln y = .... somethingtodo with (x)...$ can you see how $x = frac 3+ln y3-ln yimplies ln y = frac 3x -3x+1$?
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

The problem within my textbook is $f(x)=$$frac3+lnx3-lnx$ I checked the answer and it is $y=$$e^frac3x-3x+1$
I've tried many times to simplify the equation after switching the variables but I don't know how to seperate the $y$ from the natural logarithm. If someone could show me the steps involved I would appreciate it.

algebra-precalculus

share|cite|improve this question

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 14 mins ago

dmtri

1,6332521

edited 14 mins ago

dmtri

1,6332521

asked 1 hour ago

ShonellShonell

111

New contributor

Shonell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

ShonellShonell

111

4 Answers
4

active

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2

$begingroup$

Well, as always you want to solve for $y$ in

$x = frac 3+ln y3-ln y$

$x(3-ln y) = (3 + ln y)$

$3x - 3 = ln y + xln y$

$ln y (x+1) = 3x-3$

$ln y = frac 3x-3x+1$

$e^ln y = e^frac 3x-3x+1$

So $y = e^frac 3x-3x+1$

I don't know how to seperate the y from the natural logarithm

Then inverse of $ln y = K$ is $y = e^ln y = e^K$. $ln x$ and $e^x$ are inverses of each other. YOu "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.4k22687

$endgroup$

add a comment | 

1

$begingroup$

Write your function in the form $$3y-3=ln(x)(y+1)$$
Can you finish?

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

1

$begingroup$

Steps:

Start with:$$y=frac3+ln x3-ln x$$

Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^ln x=e^g(y)$$

Now switch $x$ and $y$ to get the final result.

share|cite|improve this answer

answered 1 hour ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

0

$begingroup$

$$y=frac3+ln x3-ln x$$

Swap $(x,y)$ to find inverse. This is the procedure!

$$x=frac3+ln y3-ln y$$

Componendo/Dividendo

$$fracx-1x+1= dfrac2 ln y6$$

$$ y= y_,(inv-fn) = e^frac3(x-1)(x+1) $$

Also the graphs mirror each other with respect to mirror line $ y=x.$

share|cite|improve this answer

edited 42 mins ago

answered 47 mins ago

NarasimhamNarasimham

21k62158

$endgroup$

add a comment | 

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2

$begingroup$

Well, as always you want to solve for $y$ in

$x = frac 3+ln y3-ln y$

$x(3-ln y) = (3 + ln y)$

$3x - 3 = ln y + xln y$

$ln y (x+1) = 3x-3$

$ln y = frac 3x-3x+1$

$e^ln y = e^frac 3x-3x+1$

So $y = e^frac 3x-3x+1$

I don't know how to seperate the y from the natural logarithm

Then inverse of $ln y = K$ is $y = e^ln y = e^K$. $ln x$ and $e^x$ are inverses of each other. YOu "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.4k22687

$endgroup$

add a comment | 

2

$begingroup$

Well, as always you want to solve for $y$ in

$x = frac 3+ln y3-ln y$

$x(3-ln y) = (3 + ln y)$

$3x - 3 = ln y + xln y$

$ln y (x+1) = 3x-3$

$ln y = frac 3x-3x+1$

$e^ln y = e^frac 3x-3x+1$

So $y = e^frac 3x-3x+1$

I don't know how to seperate the y from the natural logarithm

Then inverse of $ln y = K$ is $y = e^ln y = e^K$. $ln x$ and $e^x$ are inverses of each other. YOu "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.4k22687

$endgroup$

add a comment | 

$begingroup$

Well, as always you want to solve for $y$ in

$x = frac 3+ln y3-ln y$

$x(3-ln y) = (3 + ln y)$

$3x - 3 = ln y + xln y$

$ln y (x+1) = 3x-3$

$ln y = frac 3x-3x+1$

$e^ln y = e^frac 3x-3x+1$

So $y = e^frac 3x-3x+1$

I don't know how to seperate the y from the natural logarithm

Then inverse of $ln y = K$ is $y = e^ln y = e^K$. $ln x$ and $e^x$ are inverses of each other. YOu "cancel" $ln y=k$ by raising $e$ to the value to get $y = e^k$, and you "cancel" $e^y = M$ by taking the natural log to get $y = ln M$.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.4k22687

$endgroup$

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.4k22687

answered 1 hour ago

fleabloodfleablood

72.4k22687

answered 1 hour ago

fleabloodfleablood

72.4k22687

1

$begingroup$

Write your function in the form $$3y-3=ln(x)(y+1)$$
Can you finish?

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

1

$begingroup$

Write your function in the form $$3y-3=ln(x)(y+1)$$
Can you finish?

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

add a comment | 

$begingroup$

Write your function in the form $$3y-3=ln(x)(y+1)$$
Can you finish?

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered 1 hour ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

1

$begingroup$

Steps:

Start with:$$y=frac3+ln x3-ln x$$

Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^ln x=e^g(y)$$

Now switch $x$ and $y$ to get the final result.

share|cite|improve this answer

answered 1 hour ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

1

$begingroup$

Steps:

Start with:$$y=frac3+ln x3-ln x$$

Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^ln x=e^g(y)$$

Now switch $x$ and $y$ to get the final result.

share|cite|improve this answer

answered 1 hour ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

$begingroup$

Steps:

Start with:$$y=frac3+ln x3-ln x$$

Rewrite the equality as: $$ln x=g(y)$$ and conclude that: $$x=e^ln x=e^g(y)$$

Now switch $x$ and $y$ to get the final result.

share|cite|improve this answer

answered 1 hour ago

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

answered 1 hour ago

drhabdrhab

103k545136

answered 1 hour ago

drhabdrhab

103k545136

answered 1 hour ago

drhabdrhab

103k545136

0

$begingroup$

$$y=frac3+ln x3-ln x$$

Swap $(x,y)$ to find inverse. This is the procedure!

$$x=frac3+ln y3-ln y$$

Componendo/Dividendo

$$fracx-1x+1= dfrac2 ln y6$$

$$ y= y_,(inv-fn) = e^frac3(x-1)(x+1) $$

Also the graphs mirror each other with respect to mirror line $ y=x.$

share|cite|improve this answer

edited 42 mins ago

answered 47 mins ago

NarasimhamNarasimham

21k62158

$endgroup$

add a comment | 

0

$begingroup$

$$y=frac3+ln x3-ln x$$

Swap $(x,y)$ to find inverse. This is the procedure!

$$x=frac3+ln y3-ln y$$

Componendo/Dividendo

$$fracx-1x+1= dfrac2 ln y6$$

$$ y= y_,(inv-fn) = e^frac3(x-1)(x+1) $$

Also the graphs mirror each other with respect to mirror line $ y=x.$

share|cite|improve this answer

edited 42 mins ago

answered 47 mins ago

NarasimhamNarasimham

21k62158

$endgroup$

add a comment | 

$begingroup$

$$y=frac3+ln x3-ln x$$

Swap $(x,y)$ to find inverse. This is the procedure!

$$x=frac3+ln y3-ln y$$

Componendo/Dividendo

$$fracx-1x+1= dfrac2 ln y6$$

$$ y= y_,(inv-fn) = e^frac3(x-1)(x+1) $$

Also the graphs mirror each other with respect to mirror line $ y=x.$

share|cite|improve this answer

edited 42 mins ago

answered 47 mins ago

NarasimhamNarasimham

21k62158

$endgroup$

share|cite|improve this answer

edited 42 mins ago

answered 47 mins ago

NarasimhamNarasimham

21k62158

answered 47 mins ago

NarasimhamNarasimham

21k62158

answered 47 mins ago

NarasimhamNarasimham

21k62158