# solve (-1)^x=1

by Apteronotus
Tags: solve
 P: 196 Ok. So the answer to finding the solution of $(-1)^x=1$ is clear. But say we didnt know it and wanted to solve it. One approach is to take the log of both sides $x\cdot log(-1)=log(1)=0$ But now the right hand side is defined where as the left is not! What am I missing?
P: 3,173
 Quote by Apteronotus 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 $\frac{x}{(x-1)} = \frac{1}{(x-1)}$ by multiplying both sides by $x-1$.

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 $f(x) = g(x)$ and do some manipulation on it to produce another equation $h(x) = r(x)$ then we are suppose to ask if the manipulation may have created an equation that has more or fewer solutions than the original equation.
 P: 498 You need a logarithm that's defined for complex numbers.
P: 196

## solve (-1)^x=1

 Quote by Stephen Tashi 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.
P: 196
 Quote by olivermsun You need a logarithm that's defined for complex numbers.
Olivermsun, I believe in complex terms we have $log(-1)=i\pi$. But again we face the same dilemma as
$i\pi\ne 0$
 P: 498 Maybe let me ask this way: what are the complex logs of 1?
P: 3,173
 Quote by Apteronotus 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 $(-1)^x = 1$ 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 $log(-1) = i \pi$ and transform the new equation to $x (i \pi) = 0$ then $x = 0$ is a solution to the transformed equation.
Mentor
P: 15,592
 Quote by Stephen Tashi Mathematical manipulations are intended as a way to abbreviate thinking, not as a way of eliminating it.
Best. Statement. Ever.
P: 1,351
 Quote by Apteronotus 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.
P: 427
 Quote by Vanadium 50 Best. Statement. Ever.
Seconded. That's going on the board tomorrow.
 P: 422 $$\\(-1)^x=1 \\ \\x=\log_{-1}(1) \\ \\x=\frac{\ln(1)}{\ln(-1)} \\ \\x=\frac{0}{i \pi} \\ \\x=0$$
 P: 5 you cannot add log because log(-1) is not defined.
 P: 5 (-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
HW Helper
P: 2,151
 Quote by Apteronotus Olivermsun, I believe in complex terms we have $log(-1)=i\pi$. But again we face the same dilemma as $i\pi\ne 0$
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
$$e^{x \, \pi \imath}=e^0 \\ \text{we conclude}\\ x \, \pi \imath=2n \, \pi \imath \\ \text{for some n \in \mathbb{Z}}$$

 Related Discussions General Math 2 Precalculus Mathematics Homework 6 Calculus 2 Academic Guidance 16 Precalculus Mathematics Homework 9