7.2 Techniques for Solving Exponential Functions

Hi, this post covers 7.2 Techniques for Solving Exponential Functions.

- Radioactive Decay
In this chapter, we learned how to find the half-life of radioactive substances, which means we know how to find the half-life of pretty much anything that works in a similar fashion. The equation we are given is:

A(t) = A₀(1/2)^t/h

All we have to do now is substitute a corresponding value in each variable.
t = time, h = half-life, A(t) = mass of substance after a period of time, and A_{0 =} mass of substance at 0 time. If you have any 3 of the 4 variables, then it is possible to isolate and solve for the unknown variable. An example is shown on page 370 of our textbooks.

- Powers With Different Bases
For these types of questions, we can either solve by Applying Algebraic Reasoning or Using a Graphing Calculator.

Applying Algebraic Reasoning:
For this method, we take the log of both sides and use the power law of logarithms.

4^2x-1 = 3^x+2

log4^2x-1 = log3^x+2

(2x - 1)log4 = (x + 2)log3

2xlog4 – log4 = xlog3 + 2log3

2xlog4 – xlog3 = 2log3 + log4

x(2log4 – log3) = 2log3 + log4

x = 2log3 + log4 / 2log4 – log3

After that, just use a calculator if you want to determine the approx. value of the expression. In this case, the answer is approximately 2.14.

- Apply the Quadratic Formula
Lastly, we learned how to solve questions like this:

2^x – 2^-x = 4

In order to use the quadratic formula to solve for this type of question, we must multiply both sides by 2^x in order to generate a suitable equation:

2^x(2^x – 2^-x) = 2^x(4)

2^x(2^x) - 2^x(2^-x) = 2^x(4)

2^2x – 2⁰ = 2^x(4)

2^2x – 1 = 2^x(4)

Then use the power of a power law to the term 2^2x:

(2^x)² – 1 = 2^x(4)

And then write in standard form:

(2^x)² – 4(2^x) – 1 = 0

Then you apply the quadratic formula, leaving you with 2 + sqrt(5) and 2 - sqrt(5).

Alternatively, you could have set k = 2^x and then solved for k which would have yielded

2^x = 2 + sqrt(5) and

2^x = 2 - sqrt(5)

Which would appear to have 2 answers but after taking the common logarithm of both sides, applying the power law of logarithms, and dividing both sides by log2, you would be left with an expression that when inputted into the calculator yields approximately 2.08. As for 2^x = 2 - sqrt(5), the answer would be less than 0, which is not possible since powers always have to be a positive number so this answer can be omitted.

That's about it. Thanks for reading.