Negative logarithm within an equation problem?

[Tangent100]

Hello,

I know that a negative logarithm is undefined.

But I am faced with an equation like this:
-1 x (-2)^(n-1) = -16777216
I divided it by -1 to give (-2)^(n-1) = 16777216
And then took logs to get (n-1) log(-2) = log 16777216

Since I can't work out the log of a negative number, what do I do?

I know that the answer is 25 if i use log (2) but I don't get that :/

 

[Mark44]

Hello,

I know that a negative logarithm is undefined.

Logarithms can be negative, as for example log(.1) = -1. What you probably meant was that you can't take the log of a negative number, assuming that we're dealing with the real-valued log function.

But I am faced with an equation like this:
-1 x (-2)^(n-1) = -16777216
I divided it by -1 to give (-2)^(n-1) = 16777216
And then took logs to get (n-1) log(-2) = log 16777216

You could make an assumption about n. In this equation, (-2)^(n-1) = 16777216, if n is an odd integer (so that n - 1 is even), then (-2)^{n - 1} will be equal to (2)^(n-1) = 16777216. I would look at two cases: one where n is odd, and the other where n is even.

Since I can't work out the log of a negative number, what do I do?

I know that the answer is 25 if i use log (2) but I don't get that :/

 

[Tangent100]

Logarithms can be negative, as for example log(.1) = -1. What you probably meant was that you can't take the log of a negative number, assuming that we're dealing with the real-valued log function.
You could make an assumption about n. In this equation, (-2)^(n-1) = 16777216, if n is an odd integer (so that n - 1 is even), then (-2)^{n - 1} will be equal to (2)^(n-1) = 16777216. I would look at two cases: one where n is odd, and the other where n is even.

Thank you for quick response.

So it works similar to the case where (-1)^even = positive and (-2)^odd = negative...
so if n-1 was odd (n was even), then the equation would be unsolvable...

It's an easier solution than I thought... It's nice to know why I turned that expression positive for all these years and got the right answer!

Edit: Yes, I did mean taking a log of a negative number.