# Negative logarithm within an equation problem?

• Tangent100
In summary, the conversation discusses the concept of negative logarithms and how they can be solved in certain cases. The equation given is -1 x (-2)^(n-1) = -16777216, and by dividing it by -1 and taking logs, the solution can be obtained by assuming n to be an odd integer. The conversation also addresses the misconception of negative logarithms and provides a simpler solution than previously thought.
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 :/

Tangent100 said:
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.
Tangent100 said:
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.
Tangent100 said:
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
Mark44 said:
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.

Last edited:

## 1. What is the purpose of using a negative logarithm in an equation?

The purpose of using a negative logarithm in an equation is to solve for the exponent or power of a number that would result in the given value. It is often used to simplify and solve complex exponential equations.

## 2. How do you solve an equation with a negative logarithm?

To solve an equation with a negative logarithm, you must first isolate the logarithmic term on one side of the equation. Then, take the inverse of the logarithmic function to both sides of the equation to eliminate the logarithm. Finally, solve for the variable using basic algebraic techniques.

## 3. Can you provide an example of solving an equation with a negative logarithm?

Yes, for example, to solve the equation log(x) = -2, we first isolate the logarithmic term by dividing both sides by -1. This gives us log(x) = 2. We then take the inverse of the logarithmic function, which is 10^x, to both sides of the equation. This results in 10^log(x) = 10^2, which simplifies to x = 100. Therefore, the solution to the equation is x = 100.

## 4. Are there any restrictions when using a negative logarithm in an equation?

Yes, when using a negative logarithm in an equation, the value inside the logarithm must be a positive number. This is because taking the logarithm of a negative number is undefined in real numbers.

## 5. How does the base of the logarithm affect the solution of an equation with a negative logarithm?

The base of the logarithm does not affect the solution of an equation with a negative logarithm. As long as the base is the same on both sides of the equation, the solution will be the same. However, using a different base may result in a slightly different form of the solution, but it will still be equivalent.

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