Using complex analysis, show the solutions for log_{−2} (−8)?

Question

I cant seem to find the solution Im looking for.

Its intuitively obvious to me that the solution (one of them) can and should be +3. After all, (−2)³ = −8.

But every route I take, mathematically, either on paper or via calculator, gives me the exact same solution. Its a weird complex value.

Testing it verifies that it is -a- correct solution, as raising −2 to the solution I keep arriving at does yield −8.

How many solutions does log_{−2} (−8) have? Because I cant seem to reach an answer of 3

log_{−2} (−8) =
ln (−8) / ln (−2) =
[ ln (−1) + ln (8) ] / [ ln (−1) + ln (2) ] =
[ π⋅i + ln (8) ] / [ π⋅i + ln (2) ]... this is the solution I keep getting

Or, in polar form
√[ ( ln² 8 + π² ) / ( ln² 2 + π² ) ] ⋅ cis [ arctan (π/ln 8) - arctan (π/ln 2) ]

The absolute value of the above solution is slightly greater than 1, but less than 2. Should not the absolute value be 3 if +3 is also a valid solution?

Answer 1

The problem you are suffering from isnt the math or technique... its the simple fact that you are forgetting about the multitudes of answers created by rotations about the complex circle.

ln (−1) inst just πi... its also 3πi and 5πi and −πi

So... when you take the natural logarithm of negative one, you need to account for this.

ln (−1) = i·π(2n + 1), where 'n' is an integer.

The solution you keep arriving at is the default solution, at n=0. Calculators evaluate this way. And its the default on paper, too, as you completely forgot about this factor.

Keep in mind that the integer represented by 'n' need not be the same value in each case where you take the 'ln' of −1. You evaluate the 'ln' of −1 twice, they can each have a different 'n' integer.

==
And to answer your second question... no, not necessarily. About the absolute values. The key is that we are dealing with imaginary values here. Like taking a trig function of an imaginary value (imaginary domain), the solution can be >1 or < −1, outside the normal real-valued range of the function. In complex analysis we learn that dealing with imaginary values isnt exactly intuitively predictable. They grow at exponential rates (cis θ = e^iθ). Not the hyperbolic trig function relationships, as well.

Answer 2

I fly under the radar around here, so I'm happy to say that I feel under-appreciated. I guess the economy has hit Yahoo points, too, huh?

Answer 3

(-2)^x= -8
so x=ln(-8)/ln(-2)

ln(-8)= ln8+i*(2k+1)pi and ln(-2) = ln(2)+i*(2k´+1)*pi where k and k´are integers and not necessarily the same
so
x= ( ln 8 +i*(2k+1)*pi)/(ln 2 +i*(2k´+1)*pi) so you get infinite values.
You should try if there exist k and k´(integers) so x=3

Answer 4

let
s= ln-8/ln-2 =(πi+ 3ln2)/(πi+ln2)
s(πi+ln2) =πi+ 3ln2
e^(sπi+sln2) =e^(πi+3ln2) --->square both sides
e^(2sπi+2sln2)=e^(2πi+6ln2)
e^(2sπi) * e^(2sln2) = e^(2πi) * e^(6ln2)
as e^2sπi =e^2πi =1
e^2sln2=e^6ln2
2sln2=6ln2
s=3