“How does one solve ** x + ln(x) = c** algebraically?” A teachers asked that on the old calculus electronic discussion group (EDG) and it got me to thinking about what a “solution” really is.

We start by teaching students how to solve linear equations; the idea is to do some arithmetic and/or algebraic operations on an equation so that you end up with *x* = some number. And that’s what students learn: do some operations on an equation so that you end up with *x* = some number.

Next come quadratic equations. You can perform a series of operations (completing the square) and end up with *x* = two numbers. And then you learn two short cuts – factoring and the quadratic formula. (Okay, maybe you learn factoring first.) These solutions often involve radicals, and therefore, do not necessarily give a recognizable number.

Next come cubics and higher degree polynomials. Sometimes you can factor, and there is a cubic formula. There is also the rational root theorem and synthetic division that can help you find rational solutions to polynomial equations; but if the solutions are not rational, you may be out of luck.

Equations involving trigonometric functions *always *have solutions that are the so-called “special angles.” Exponential and logarithmic equation *always* involve convenient bases. So everything can be solved that way – or not.

These kinds of answers are usually called “closed form” expressions. That is, they are given in a notation that indicates what arithmetic must be done to get a decimal answer. That is, 5192/13 means to get “the answer” divide 5192 by 13. Since decimals are not always “exact,” the closed-form answers are preferred – they are “exact.”

We rarely consider solving by graphing by-hand, graphing calculator, computer algebra systems (CAS), or searching the internet. Why not? Because by-hand graphing is not very accurate, you only get numerical answer, graphing is “just for checking,” and of course technology is not allowed on the state exams.

So returning to the EDG here are some of the answers that were offered to the question “How does one solve ** x + ln(x) = c** algebraically?”

- You don’t
- You can’t
- Sure you can like this:
- Where Lambert(
*W*) is the Lambert*W*-function named after Johann Heinrick Lambert (1728 – 1777).

Then folks started complaining.

Someone wrote “’Naming’ a previously unknown functions isn’t ‘solving’ a problem.”

To which someone else replied

- Before Joe Arcsine invented his function, one could not “solve” sin(x) = 0.123
- Before Betty Logarithm invented her function, one could not “solve”
- Before Johann Lambert invented his function, one could not “solve”

And he is correct – well okay, his idea is correct, even if he may not have ascribed the solutions to the correct mathematicians. We actually do it all the time. Functions like the arcsin(*x*), ln(*x*), and the like are simply names given to function for which we may not have a series of arithmetical operations leading to a closed-form solution. (Taylor series, being infinite in length, are not closed-form.)

**Simplifying**

Then there are the answers themselves. Simplifying makes things easier.

*x*= 3,*x*= 0.125, are fine, but*x*= 4.567… not so much.*x*= ½ and*x*= are great, but*x*= is not.- We like
*x*= , we like , we’re not too sure about*x*= ,*x*= is just plain wrong, but*x*= is okay. *x*= cannot be correct because we have to rationalize the denominator; try that with*x*=- Likewise,
*x*= is okay, but*x*= is not. - Which of
*x*= and the equivalent*x*= is simplified? - When we differentiate tan(
*x*) using the quotient rule we get , but that has to be changed to even though there is no secant button on a calculator. - And then there is Lambert
*W*(e^{4})

So what have we learned?

- Each kind of equation, has a different process for finding its solution.
- We are allowed to make up new functions to solve equations.
- To the outsider (read: student) this looks like a hodgepodge – and for good reason.
- Most of our “solutions” are really directions for finding the solution.

Of course solving y =x^2 by x = sqrt(y) is not different from solving y=x exp(x) by x = W(y).

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