Project 11: Solving Exponential Equations

Intro Problem:  A piece of paper is .01 inches thick.  If we were to fold it in half, it would be .02 inches thick.  If we were to fold in half again, it would be .04 inches thick.

How many folds would we need to make for the paper to be 1.28 inches thick?

The thickness of our paper can be modeled by: Y = .01(2)^x  where x = the number of folds and y = the thickness

Therefore, we can solve .01(2)^x = 1.28

Simplifying we get, 2^x = 128

This equation is fairly simple since we can keep multiplying by 2 until we get to 128.  However, some problems require more work.

We could take the log of both sides:

log(2^x) = log(128)

A property of logarithms shows we can rewrite as:

X*log2 = log128

X = log128/log2 = 7

So we would need to fold the paper 7 times for the thickness to be 1.28 inches.

Project 11: Variables 'a',' b', and 'c' have been initialized. There is a working method called log(a)

These coefficients represent the value in an exponential equation as shown above (3, 2, -24 in the first example)

log(x) returns log base 10 of x

example: log(100) returns 2

Task: Appropriately initialize the value of 'solution' that represents the solution to the equation

Note: You can assume that the exponential equation will have a solution.

**If your code works for 5 test cases in a row, you can enter your e-mail address.

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0