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Is a negative logarithm meaningless?

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1

$begingroup$

Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt-a$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.

For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?

So can you evaluate negative logarithms?

real-analysis complex-analysis logarithms

share|cite|improve this question

asked Mar 12 at 21:40

GurjinderGurjinder

552417

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^i pi$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
  $endgroup$
  – Ryan Goulden
  Mar 12 at 21:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in complex plane yes.
  $endgroup$
  – cand
  Mar 12 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
  $endgroup$
  – Peter
  Mar 12 at 22:38
  

  
  
  
  

add a comment | 

1

$begingroup$

Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt-a$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.

For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?

So can you evaluate negative logarithms?

real-analysis complex-analysis logarithms

share|cite|improve this question

asked Mar 12 at 21:40

GurjinderGurjinder

552417

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You absolutely can, and the extension is due to complex analysis. Assuming the principle branch of $log z$, consider $z=-1=e^i pi$ and so $log z = i pi$. In general, $log z = ln|z| + i pi$
  $endgroup$
  – Ryan Goulden
  Mar 12 at 21:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in complex plane yes.
  $endgroup$
  – cand
  Mar 12 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It depends whether you allow complex numbers, but even then, the logarithms of negative numbers are not unique, you have to choose a particular branch to make it unique.
  $endgroup$
  – Peter
  Mar 12 at 22:38
  

  
  
  
  

add a comment | 

$begingroup$

Is $log(-x)$, where $x in (0, infty)$ undefined? From solving quadratics I was first told that the discriminant has to be non-negative, since $sqrt-a$, where $a in (0, infty)$ is undefined. But this was before learning about Imaginary numbers and the complex plane.

For example, what would $log_3(-9)$ be? Is there a number $a$ such that $3^a=-9$?

So can you evaluate negative logarithms?

real-analysis complex-analysis logarithms

share|cite|improve this question

asked Mar 12 at 21:40

GurjinderGurjinder

552417

$endgroup$

share|cite|improve this question

asked Mar 12 at 21:40

GurjinderGurjinder

552417

share|cite|improve this question

asked Mar 12 at 21:40

GurjinderGurjinder

552417

asked Mar 12 at 21:40

GurjinderGurjinder

552417

asked Mar 12 at 21:40

GurjinderGurjinder

552417

1 Answer
1

active

oldest

votes

2

$begingroup$

Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrtx^2+y^2$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).

Equipped with this knowledge and the fact that
$log_a z=fracln zln a$:
$$
log_3(-9)=fracln9+ipiln3=2+fracpi iln3.
$$

share|cite|improve this answer

answered Mar 12 at 22:36

useruser

5,41411030

$endgroup$

add a comment | 

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2

$begingroup$

Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrtx^2+y^2$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).

Equipped with this knowledge and the fact that
$log_a z=fracln zln a$:
$$
log_3(-9)=fracln9+ipiln3=2+fracpi iln3.
$$

share|cite|improve this answer

answered Mar 12 at 22:36

useruser

5,41411030

$endgroup$

add a comment | 

2

$begingroup$

Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrtx^2+y^2$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).

Equipped with this knowledge and the fact that
$log_a z=fracln zln a$:
$$
log_3(-9)=fracln9+ipiln3=2+fracpi iln3.
$$

share|cite|improve this answer

answered Mar 12 at 22:36

useruser

5,41411030

$endgroup$

add a comment | 

$begingroup$

Yes, it is possible to evaluate logarithm of a negative number in the complex plane. Moreover it is possible evaluate logarithm of any complex number $z=x+iy$:
$$
log z=ln|z|+iarg(z).
$$
where the real numbers $|z|=sqrtx^2+y^2$ and $arg z$ are, respectively, the absolute value and argument of $z$. The argument is essentially the angle in the complex plane between $z$ and positive direction of the real axis. There is however a complication. Different from the real logarithm the complex one is multivalued function, so that any multiple of $2pi i$ can be added to its value. One of possible solution of the problem is to bound the imaginary part (for example from $-pi$ to $pi$).

Equipped with this knowledge and the fact that
$log_a z=fracln zln a$:
$$
log_3(-9)=fracln9+ipiln3=2+fracpi iln3.
$$

share|cite|improve this answer

answered Mar 12 at 22:36

useruser

5,41411030

$endgroup$

share|cite|improve this answer

answered Mar 12 at 22:36

useruser

5,41411030

answered Mar 12 at 22:36

useruser

5,41411030

answered Mar 12 at 22:36

useruser

5,41411030