-1^x = b is this solveable using natural logarithms?



-1^x = b
[Multiply both sides by -1]
1^x= -b
x • ln(1) = ln(-b)
x • 0 = ln(-b)
0 = ln(-b)

Where can you go, if anywhere, from here?

Reply 1

0gg

Multiplying by -1 in the first step gives -1^(x+1) = -b.

-1^x (when x is an integer) will give -1 or 1, so if b is not either of those (given b and x are real), there aren't any solutions.

There's probably some complex numbers wacky stuff possible here though that someone more knowledgeable could figure out 🤔

Reply 2

tonyiptony

Original post by 0gg

Multiplying by -1 in the first step gives -1^(x+1) = -b.
-1^x (when x is an integer) will give -1 or 1, so if b is not either of those (given b and x are real), there aren't any solutions.
There's probably some complex numbers wacky stuff possible here though that someone more knowledgeable could figure out 🤔

In complex world, taking logarithm is a bit tricky. In fact we use a different notation for complex logarithm - Log(z) (not sure how universal this is). The dreadful notation of ln(x) is only reserved for the reals, and ln(-1) is undefined.

Without going into too much detail, Log(-1) happens to also be undefined, despite you could take Log of almost all complex numbers.

Little log(z) gives you infinitely many solutions. Again, details omit here. But the overarching handwavy idea is little log(z) is not a function in the complex world, so it takes a lot more care. My one semester of complex analysis tells me to run away from it.

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Anyhow that's what I've gathered so far, if I've learned things right.

But back to the original question, the troubling line is line 3 (or 4) - you are only allowed to take log if b is negative (and as such, -b is positive).

(edited 1 year ago)

Reply 3

mqb2766

Original post by Αbsolute


-1^x = b
[Multiply both sides by -1]
1^x= -b
x • ln(1) = ln(-b)
x • 0 = ln(-b)
0 = ln(-b)

Where can you go, if anywhere, from here?

Just to expand on Ogg's comments for the real valued case
(-1)^x = b
only has (real) solutions if b=+/-1 and in those cases, the solutions are the even and odd integers respectively. So when there is a solution, there are an infinite number of solutions and this is related to your x*ln(1) =x*0. For any other value of b, there are no real solutions. As Ogg noted as well (-1)^x is different from -(1^x). Writing it as -1^x is the latter using standard precedence operations, whereas you probably meant the former? So your line 1-2 may not be correct.

Your working would be correct for
-(1^x) = -1
so you must have b=-1 and taking logs youd end up with
x*0 = 0
which is indeed correct as its satisfied for all real x.