Solving X^2 = 2^X ```Name: Matthew. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` Question: Question: Hi there, last year my precalculus teacher offered me a way out of his final (which happens to be the most feared final in my high school). What happened is that he made a claim and I bet that he was wrong. After posing the problem of solving X^2 = 2^X he showed us how easy it was to solve graphically. All we had to do was graph x^2 on the calculator with 2^x and have the calculator numerically fine where the two intersected. The solutions are 2,4, and a negative number (which i believe to be irrational) that the calculator cuts off at -.7666647. He uses this example to show parents the necessity of Graphing calculators, and while I agree that they are fairly necessary for the course I find it impossible to believe that there is no algebraic way to solve that equation. I spent almost half a year trying different things, different manipulations, log rules, I even did some research on John Napier as an attempt to calculate log base 2 on the off chance that would help.... the closest I came was an attempt at a Taylor series (even though at the time I did not properly understand derivatives) but the polynomials that I created that modeled it closely enough were too difficult to solve for. Long story short, is there a way to solve X^2=2^X Algebraically... or any way other than a graphing calculator. Replies: Matthew, Using precalculus techniques, you can show that x^2 = 2^x has the same positive roots as kx = ln(x), where k = ln(2)/2. As you noted, the solutions of this latter equation are 2 and 4 (by trial and error). For arbitrary positive constant, c, the equation cx = ln(x) has either 0, 1 or 2 roots depending on whether c > 1/k, c = 1/k or c > 1/k. (Use a slope argument with calculus.). To get the negative root, use symmetry and the equation x^2 = (1/2)^x, which has the same positive root as -kx = ln(x). This last equation evidently has only one root, the opposite of which is the negative root of the original equation. So, it comes down to asking when the root of -cx = ln(x) has algebraic solutions. Start with -c = ln(a). Then ln(a)x=ln(x) Paul Beem The equation: X^2 = 2^X is not algebraic because the variable 'X' appears as an exponent. So, for example, the properties of algebraic polynomial equations of degree 'n' (the highest power having 'n' solutions) no longer has meaning because the degree of the equation varies with the choice of 'X'. The solution you seek is a transcendental number (-0.7666646959...) -- a few more decimal places than your calculator could handle, but still having no repeating decimal strings. It belongs to a class of equations "iterated exponential constants". If you wish to pursue this, you can start with the book "Mathematical Constants" section [6.11] page 448 by Steven R. Finch. Now for the "soap box" lecture. Given enough time, patience, and computer memory, you can slug out a solution numerically -- but unless you need "the value" of the number to build a bridge -- that does not give you any insight about the mathematical concepts. So graphing calculators are OK for "crunching numbers" but they provide no understanding. There are any number of numerical methods for computing the value of a function to any desired degree of accuracy -- but they provide no understanding of how and why that particular number exists. The seemingly simple equation is just the tip of an iceberg of some rather advanced mathematical analysis. Vince Calder Click here to return to the Mathematics Archives

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