Solving unknown negative exponents?

Question

I know I posted earlier but I forgot to make the exponents negative. Here is my question again, how to solve for x:

200 = 4.3222^-x + 4.3222^-2x

Answer 1

This problem doesn't seem to be solvable by normal algebraic methods. Logarithms are pretty much out since the of the LHS addition. So it seems numerical analysis is needed. Time to pull out newtons method.

To apply Newton's method we need to define a few thing.

f(n) = 4.3222^-x + 4.3222^-2x - 200

f'(n) = 4.3222^-x * ln 4.3222 * -1 + 4.3222^-2x * ln 4.3222 * -2 + 0

Next we need an approximation of the location of the zero(s) of f(n). We know that as x increases both 4.3222^-x and 4.3222^-2x will rapidly approach zero. As x decreases they will rapidly increase. Since 4.3 is around e let's make our guess to be around -.5 *ln(200)/ln(4.3222) (aka .5 * the base 4.3222 log of 200). Or x is about -1.8, plugging back into the function tells us this approximation is pretty close.

Thus we get for the method that, x sub 1 = -1.8.

Newtons method is

x sub 2 = x2 = x1-f(x1)/f'(x1)

x2 = -1.8 - [4.3222^-(-1.8)+4.3222^(-2*-1.8)-200]/[4... 4.3222*-1+4.3222^(-2*-1.8)*ln4.3222*-2+0...

x2 = -1.7859680754933350628405769965494

repeating until we get enough precision provides us with a 32 digit answer of

x6 = -1.7856770326310667491865644256693

Final answer n = -1.785677 (remember to check your answer with a calculator) Further proof to check that this is the only root can be done by realizing that the first derivative of the function is negative for all x. Thus the original function can come back to y=0.