Solve for n How can we use logarithms to solve for the variable in this equation?

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In summary, the equation 2n+2=0 has no solution if n is a real number, but has solutions if n can be complex. To solve for complex solutions, use the polar form of complex numbers and take the ln of both sides to start.
  • #1
vcsharp2003
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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
 
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  • #2
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.
 
  • #3
eys_physics said:
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?
 
  • #4
vcsharp2003 said:
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.
 
  • #5
eys_physics said:
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 e2kπi where k is any integer
and -1 = eπi.

I am not sure what value of k would I take when expressing 2 in polar form since k could be any integer?
 
  • #6
eys_physics said:
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 (2eπi)
n ln2 = ln2 + ln(eπi)
n ln2 = ln2 + iπ lne
n ln2 = ln2 + iπ
n = 1 + i (π/ln2)
 
Last edited:

1. What does "solve for n" mean?

"Solve for n" means to find the value of the variable n in an equation or expression. It is typically used in algebraic or mathematical contexts.

2. How do I solve for n?

To solve for n, you need to isolate the variable on one side of the equation and perform the necessary operations to get n by itself. This may involve using mathematical properties such as addition, subtraction, multiplication, or division.

3. What is the purpose of solving for n?

Solving for n allows you to find the specific value of the variable in an equation or expression. This can help you solve problems, make predictions, or understand patterns in the data.

4. Can you give an example of solving for n?

Sure, let's say we have the equation 2n + 5 = 17. To solve for n, we need to get n by itself on one side of the equation. First, we subtract 5 from both sides to get 2n = 12. Then, we divide both sides by 2 to get n = 6. Therefore, the value of n in this equation is 6.

5. Is it always possible to solve for n?

In most cases, yes. However, there may be some equations or expressions that do not have a solution for n or have an infinite number of solutions. Additionally, some equations may have complex or imaginary solutions for n. It is important to check your work and make sure your solution makes sense in the context of the problem.

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