Solving (-1)^x=1: What Am I Missing?

  • Thread starter Apteronotus
  • Start date
Z}$}$$In summary, the conversation discusses the solution to the equation (-1)^x = 1 and the use of logarithms to solve it. However, it is pointed out that the operation of taking the logarithm of both sides is not always valid and can produce an equation with a different solution set. It is also mentioned that the complex logarithm can be used to solve the equation, but it still results in a solution set that does not match the original equation. Ultimately, it is concluded that the only solution to the equation is when x is an even integer.
  • #1
Apteronotus
202
0
Ok. So the answer to finding the solution of
[itex](-1)^x=1[/itex]
is clear.

But say we didnt know it and wanted to solve it. One approach is to take the log of both sides

[itex]x\cdot log(-1)=log(1)=0[/itex]

But now the right hand side is defined where as the left is not!

What am I missing?
 
Mathematics news on Phys.org
  • #2
Apteronotus said:
What am I missing?

You're missing the fact that there is no mathematical principle that says you can always solve an equation by "doing the same thing to both sides".

For example, you can't solve the equation [itex] \frac{x}{(x-1)} = \frac{1}{(x-1)} [/itex] by multiplying both sides by [itex] x-1 [/itex].

Mathematical manipulations are intended as a way to abbreviate thinking, not as a way of eliminating it.

When we have an equation of the form [itex] f(x) = g(x) [/itex] and do some manipulation on it to produce another equation [itex] h(x) = r(x) [/itex] then we are suppose to ask if the manipulation may have created an equation that has more or fewer solutions than the original equation.
 
  • Like
Likes 1 person
  • #3
You need a logarithm that's defined for complex numbers. :wink:
 
  • #4
Stephen Tashi said:
You're missing the fact that there is no mathematical principle that says you can always solve an equation by "doing the same thing to both sides".

You're missing the point. It's not a matter of solving the equation, its the fact that by performing a completely legitimate operation we now have an equation where the RHS does not equal the LHS.
 
  • #5
olivermsun said:
You need a logarithm that's defined for complex numbers. :wink:

Olivermsun, I believe in complex terms we have [itex]log(-1)=i\pi[/itex]. But again we face the same dilemma as
[itex]i\pi\ne 0[/itex]
 
  • #6
Maybe let me ask this way: what are the complex logs of 1?
 
  • #7
Apteronotus said:
You're missing the point. It's not a matter of solving the equation, its the fact that by performing a completely legitimate operation we now have an equation where the RHS does not equal the LHS.

What do you mean by "a completely legitimate operation"?

You aren't using the proper terminology for talking about equations. An equation in a variable x is a statement that two functions of x are equal. A solution to the equation is a value of x that makes the statement true. The set of all solutions to the equation is called "the solution set". The left hand side of an equation may not equal the right hand side of the equation for some , or for any values of the variable x.

I think what you are trying to say is that taking the log of both sides of the equation [itex] (-1)^x = 1 [/itex] transforms it to another equation, which has a different solution set (namely the null set since the transformed equation has no solutions).

There is no mathematical principle that says applying the natural logarithm function to both sides of an equation will transform it to a new equation that has the same solution set as the original equation.

If you define a complex logarithm such that [itex] log(-1) = i \pi [/itex] and transform the new equation to [itex] x (i \pi) = 0 [/itex] then [itex] x = 0 [/itex] is a solution to the transformed equation.
 
  • #8
Stephen Tashi said:
Mathematical manipulations are intended as a way to abbreviate thinking, not as a way of eliminating it.

Best. Statement. Ever.
 
  • #9
Apteronotus said:
You're missing the point. It's not a matter of solving the equation, its the fact that by performing a completely legitimate operation we now have an equation where the RHS does not equal the LHS.

For real numbers a and b, log (a^b) = a log(b) is not valid if a<0.
 
  • #10
Vanadium 50 said:
Best. Statement. Ever.

Seconded. That's going on the board tomorrow.
 
  • #11
[tex]\\(-1)^x=1 \\ \\x=\log_{-1}(1) \\ \\x=\frac{\ln(1)}{\ln(-1)} \\ \\x=\frac{0}{i \pi} \\ \\x=0[/tex]
 
  • #12
you cannot add log because log(-1) is not defined.
 
  • #13
(-1)^2y = (-1)^ ( even integer) = 1

(-2)^(2y+1) = (-1)^(odd integer) = -1

since 2y and (2y+1) form all N ( natural number) then you can say the only solution will be if x is even, i.e x= 2y , and y belongs to N
 
  • #14
Apteronotus said:
Olivermsun, I believe in complex terms we have [itex]log(-1)=i\pi[/itex]. But again we face the same dilemma as
[itex]i\pi\ne 0[/itex]
consider the equivalent equation
$$e^{x \, \log(-1)}=e^0
\\
e^{x \, \pi \imath}=e^0$$

In general we cannot conclude
x=y
from
f(x)=f(y)

In this case exp is a periodic function with period 2π i

so from
$$e^{x \, \pi \imath}=e^0
\\ \text{we conclude}\\
x \, \pi \imath=2n \, \pi \imath
\\ \text{for some $n \in \mathbb{Z}$} $$
 

1. What does (-1)^x=1 mean?

This notation is called an exponential equation, where x is the variable and -1 is the base. It means that when -1 is raised to the power of x, the result is equal to 1.

2. How do you solve (-1)^x=1?

To solve this equation, we can take the logarithm of both sides. This will give us x = 0, since log base -1 of 1 is equal to 0. Alternatively, we can also rewrite the equation as (-1)^x= (-1)^0 and use the property of exponents to get x = 0.

3. Why is x = 0 the only solution?

When solving exponential equations, we must consider the domain of the base. In this case, the base is -1, which has a domain of all real numbers except 0. Since we cannot raise -1 to any power to get 1, x = 0 is the only solution that satisfies the equation.

4. Can (-1)^x=1 have complex solutions?

No, this equation only has real solutions. Complex solutions involve imaginary numbers, and since the base -1 is a real number, the solutions must also be real.

5. What are some real-life applications of solving (-1)^x=1?

Exponential equations are commonly used in various fields such as physics, engineering, and finance. Specifically, the equation (-1)^x=1 can represent situations involving growth or decay, such as population growth or radioactive decay.

Similar threads

  • General Math
Replies
13
Views
1K
  • General Math
Replies
3
Views
1K
  • General Math
Replies
1
Views
648
  • General Math
Replies
13
Views
2K
  • General Math
Replies
5
Views
2K
  • General Math
Replies
2
Views
704
Replies
14
Views
1K
  • General Math
Replies
15
Views
2K
  • General Math
Replies
22
Views
402
Replies
2
Views
1K
Back
Top