How Can Logarithms Solve for n in This Equation?

[vcsharp2003]

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

 

[eys_physics]

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.

 

[vcsharp2003]

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?

 

[eys_physics]

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.

 

[vcsharp2003]

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 e^2kπi where k is any integer
and -1 = e^πi.

I am not sure what value of k would I take when expressing 2 in polar form since k could be any integer?

 

[vcsharp2003]

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 (2e^πi)
n ln2 = ln2 + ln(e^πi)
n ln2 = ln2 + iπ lne
n ln2 = ln2 + iπ
n = 1 + i (π/ln2)