How Can Logarithms Solve for n in This Equation?

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Discussion Overview

The discussion revolves around solving the equation 2n + 2 = 0 for the variable n. Participants explore the implications of n being a real or complex number, and the methods for finding solutions, particularly through the use of logarithms and polar forms of complex numbers.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that if n is restricted to real numbers, the equation has no solution, as 2n is always positive.
  • Others propose that if n can be complex, solutions do exist for the equation.
  • There is a suggestion to rearrange the equation to 2^n = -2 or 2^{n-1} = -1 to explore complex solutions.
  • Participants discuss expressing both 2 and -1 in polar form, noting that 2 can be expressed as 2e^{2kπi} for any integer k, while -1 is expressed as e^{πi}.
  • One participant raises uncertainty about the choice of k when expressing 2 in polar form, indicating that k could be any integer.
  • Another participant suggests taking the natural logarithm of both sides of the original equation as a method to solve for n, leading to a derived expression for n involving complex numbers.

Areas of Agreement / Disagreement

Participants generally agree that the equation has no solution for real numbers, but there is a divergence of views on how to approach finding complex solutions, with multiple methods proposed and no consensus on the best approach.

Contextual Notes

The discussion includes assumptions about the nature of n (real vs. complex) and the implications of using logarithms and polar forms, which may not be universally accepted or resolved.

vcsharp2003
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How do I go about solving the following equation for n? From inspection, it seems that this equation is not possible since 2n is always positive so positive + 2= 0 is impossible.

2n+ 2 = 0
 
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If ##n## is supposed to be a real number (or integer) the equation has no solution. But, if ##n## can be complex the equation has solutions.
 
eys_physics said:
If ##n## is supposed to be a real number (or integer) the equation has no solution. But, if ##n## can be complex the equation has solutions.

Ok that makes sense. It is not mentioned that n is a real number. For complex solutions how would I start solving this?
 
vcsharp2003 said:
Ok that makes sense. It is not mentioned that n is a real number. For complex solutions how would I start solving this?
Rearranging the equation, you have
##2^n=-2## or
##2^{n-1}=-1##
Use then the polar form ##z=re^{i\theta}## of a complex number.
 
eys_physics said:
Rearranging the equation, you have
##2^n=-2## or
##2^{n-1}=-1##
Use then the polar form ##z=re^{i\theta}## of a complex number.
So, I would express 2 in polar form and also -1 in polar form.
∴ 2 = 2 e2kπi where k is any integer
and -1 = eπi.

I am not sure what value of k would I take when expressing 2 in polar form since k could be any integer?
 
eys_physics said:
Rearranging the equation, you have
##2^n=-2## or
##2^{n-1}=-1##
Use then the polar form ##z=re^{i\theta}## of a complex number.
I think it might be easier to take original equation so we have 2n= -2, then take ln of both sides and express only -2 in polar form.

Another important idea relating to solving for an unknown in an equation is that when the unknown variable appears in a power, then taking log of both sides can help as a first step to the solution. In this problem, the unknown n appears in the power, therefore it would be a good idea to take log of both sides as a first step, which is what is done below.

ln(2n) = ln(-2)
n ln2 = ln (2eπi)
n ln2 = ln2 + ln(eπi)
n ln2 = ln2 + iπ lne
n ln2 = ln2 + iπ
n = 1 + i (π/ln2)
 
Last edited:

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