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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
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Update: June 2012
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