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Range of a log function

0

1

$begingroup$

Find the range of $f(x)=ln(fracsqrt8-x^2x-2)$

My attempt :-
$$f(x)=ln(fracsqrt8-x^2x-2)$$
$$=0.5ln(8-x^2)-ln(x-2)$$

After this I can't solve and not able to find the range.

algebra-precalculus functions logarithms

share|cite|improve this question

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Over what domain is $ln$ defined?
  $endgroup$
  – Brian
  Apr 9 at 3:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint $:$ The domain of $f$ is $(2,2sqrt 2).$
  $endgroup$
  – Dbchatto67
  Apr 9 at 3:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A possible approach could be trying to find the range of $g(x)=fracsqrt8-x^2x-2$ first. And then to check how it is transformed by the logarithm.
  $endgroup$
  – Martin Sleziak
  Apr 9 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find the domain but how to find the range using the domain
  $endgroup$
  – Abhishek Kumar
  Apr 9 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AbhishekKumar what values does $f$ take at the endpoints? Also $f$ is continuous on its domain...
  $endgroup$
  – gt6989b
  Apr 9 at 3:50
  
  

  
  
  
  

 | 
show 1 more comment

0

1

$begingroup$

Find the range of $f(x)=ln(fracsqrt8-x^2x-2)$

My attempt :-
$$f(x)=ln(fracsqrt8-x^2x-2)$$
$$=0.5ln(8-x^2)-ln(x-2)$$

After this I can't solve and not able to find the range.

algebra-precalculus functions logarithms

share|cite|improve this question

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Over what domain is $ln$ defined?
  $endgroup$
  – Brian
  Apr 9 at 3:44
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint $:$ The domain of $f$ is $(2,2sqrt 2).$
  $endgroup$
  – Dbchatto67
  Apr 9 at 3:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A possible approach could be trying to find the range of $g(x)=fracsqrt8-x^2x-2$ first. And then to check how it is transformed by the logarithm.
  $endgroup$
  – Martin Sleziak
  Apr 9 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I find the domain but how to find the range using the domain
  $endgroup$
  – Abhishek Kumar
  Apr 9 at 3:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AbhishekKumar what values does $f$ take at the endpoints? Also $f$ is continuous on its domain...
  $endgroup$
  – gt6989b
  Apr 9 at 3:50
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Find the range of $f(x)=ln(fracsqrt8-x^2x-2)$

My attempt :-
$$f(x)=ln(fracsqrt8-x^2x-2)$$
$$=0.5ln(8-x^2)-ln(x-2)$$

After this I can't solve and not able to find the range.

algebra-precalculus functions logarithms

share|cite|improve this question

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

$endgroup$

share|cite|improve this question

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

share|cite|improve this question

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

edited Apr 9 at 3:43

Martin Sleziak

45k10123277

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

asked Apr 9 at 3:41

Abhishek KumarAbhishek Kumar

867

1 Answer
1

active

oldest

votes

0

$begingroup$

Hint:

Let $dfrac8-x^2(x-2)^2=y$

$iff x^2(y+1)-4x+4y-8=0$

$x=dfrac2pmsqrt52-(4y-2)^22=2pmsqrt13-(2y-1)^2$

Now as $2<x<2sqrt2$ and $2-sqrt13-(2y-1)^2le2$

$x=2+sqrt13-(2y-1)^2$

$iff2<2+sqrt13-(2y-1)^2<2sqrt2$

$2<2+sqrt13-(2y-1)^2implies13-(2y-1)^2ne0iff(2y-1)^2<13$

$2+sqrt13-(2y-1)^2<2sqrt2implies13-(2y-1)^2<(2sqrt2-2)^2$

share|cite|improve this answer

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

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0

$begingroup$

Hint:

Let $dfrac8-x^2(x-2)^2=y$

$iff x^2(y+1)-4x+4y-8=0$

$x=dfrac2pmsqrt52-(4y-2)^22=2pmsqrt13-(2y-1)^2$

Now as $2<x<2sqrt2$ and $2-sqrt13-(2y-1)^2le2$

$x=2+sqrt13-(2y-1)^2$

$iff2<2+sqrt13-(2y-1)^2<2sqrt2$

$2<2+sqrt13-(2y-1)^2implies13-(2y-1)^2ne0iff(2y-1)^2<13$

$2+sqrt13-(2y-1)^2<2sqrt2implies13-(2y-1)^2<(2sqrt2-2)^2$

share|cite|improve this answer

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

0

$begingroup$

Hint:

Let $dfrac8-x^2(x-2)^2=y$

$iff x^2(y+1)-4x+4y-8=0$

$x=dfrac2pmsqrt52-(4y-2)^22=2pmsqrt13-(2y-1)^2$

Now as $2<x<2sqrt2$ and $2-sqrt13-(2y-1)^2le2$

$x=2+sqrt13-(2y-1)^2$

$iff2<2+sqrt13-(2y-1)^2<2sqrt2$

$2<2+sqrt13-(2y-1)^2implies13-(2y-1)^2ne0iff(2y-1)^2<13$

$2+sqrt13-(2y-1)^2<2sqrt2implies13-(2y-1)^2<(2sqrt2-2)^2$

share|cite|improve this answer

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

$begingroup$

Hint:

Let $dfrac8-x^2(x-2)^2=y$

$iff x^2(y+1)-4x+4y-8=0$

$x=dfrac2pmsqrt52-(4y-2)^22=2pmsqrt13-(2y-1)^2$

Now as $2<x<2sqrt2$ and $2-sqrt13-(2y-1)^2le2$

$x=2+sqrt13-(2y-1)^2$

$iff2<2+sqrt13-(2y-1)^2<2sqrt2$

$2<2+sqrt13-(2y-1)^2implies13-(2y-1)^2ne0iff(2y-1)^2<13$

$2+sqrt13-(2y-1)^2<2sqrt2implies13-(2y-1)^2<(2sqrt2-2)^2$

share|cite|improve this answer

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

share|cite|improve this answer

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279

answered Apr 9 at 6:33

lab bhattacharjeelab bhattacharjee

229k15159279