Undergrad Can single variable equations always be solved using basic math principles?

[Fascheue]

Can all equations with a single variable be solved?

5 = (x)/(1+x^7)

How can something like this be solved for example. Sometimes I have equations with one variable and I can’t see any way to solve them with the knowledge of math that I currently have. Is it just that a higher level of math is required to solves problems like the one above or am I missing something?

 

[fresh_42]

Can all equations with a single variable be solved?

No. $y=xe^x$ for example can't.

5 = (x)/(1+x^7)

Not sure about this one. Polynomials up to degree four can be solved in a closed form. Equations of higher degrees like this one can only sometimes be solved, depending on the given case. I haven't checked this one though.

How can something like this be solved for example.

Numerically by algorithms like the Newton-Raphson method.

Sometimes I have equations with one variable and I can’t see any way to solve them with the knowledge of math that I currently have. Is it just that a higher level of math is required to solves problems like the one above or am I missing something?

It is a bit of high level mathematics (Galois theory) to see why from degree five on, polynomials cannot be generally solved anymore, and it is not quite trivial to check a certain polynomial.

 

[phyzguy]

It depends what you mean by "solved". If you mean, "write down the answer in terms of elementary functions", then the answer is no, as fresh_42 said. But let's take his example of y=x \exp(x). The solution is x=W_n(y), where Wn is the ProductLog function. This is a tabulated function where you can look up the values, so in this sense it is no different from sin(x). So do you consider the problem "solved"? If by "solved" you mean "find the number(s) for which the equation is satisfied to a sufficient degreee of accuracy", then the solutions for any equation of this type can be solved by numerical methods like Newton-Raphson, if a solution exists.

 

[FactChecker]

Are you allowing complex numbers? Complex numbers allow solutions that are not real. Any polynomial can be factored to factors of order 1 and 2 with real numbers. The order 1 factors give a solution and order 2 factors give 2 conjugate complex solutions. Your example can be manipulated to an equivalent polynomial problem, 5 (1 + x⁷) - x = 0, with x=-1 (and some complex numbers) disallowed.