What does it mean for something to grow or decay exponentially? The idea is that the independent variable is found in the exponent rather than the base. For example, comparing \(f(t)=t^2\) and \(g(t)=2^t\), notice that \(t\) is in the exponent of the \(g(t)\), so \(g(t)\) is considered an example of exponential growth but \(f(t)\) is not (since \(t\) is not in the exponent).
The general form of an exponential growth equation is \(y = a(b^t)\) or \(y=a(1+r)^t\). These equations are the same when \(b=1+r\), so our discussion will center around \(y = a(b^t)\) and you can easily extend your understanding to the second equation if you need to.
When \(b > 1\), we call the equation an exponential growth equation.
When \(b < 1\), it is called exponential decay.
When \(b = 1\), we have \(y=a(1^t)=a\), which is just linear equation and it is not considered an exponential equation.
So it is easy to see that we have the same equation in both cases and the value of \(b\) tells us if the exponential is growing or decaying. Let's watch a quick video before we go on.

A specific type of exponential growth is when \(b=e^r\) and \(r\) is called the growth/decay rate. The equation comes from the idea that the rate of change is proportional to the quantity that currently exists. This produces the autonomous differential equation

where \(k\) is a constant called the growth/decay constant/rate.
When \(k < 0\), we use the term exponential decay.
When \(k > 0\), we use the term exponential growth.
To solve this differential equation, there are several techniques available to us. We will use separation of variables. Although not explicitly written in this equation, the independent variable we usually use in these types of equations is t to represent time.

\(\displaystyle{ \begin{array}{rcl} y' & = & ky \\ \frac{dy}{dt} & = & ky \\ \frac{dy}{y} & = & k~dt \\ \int{ \frac{dy}{y} } & = & \int{ k~dt } \\ \ln|y| & = & kt+C \\ e^{\ln|y|} & = & e^{kt+C} \\ y & = & Ae^{kt} \end{array} }\)

The constant A is usually called the initial amount since at time \(t=0\), we have \(y = Ae^0 = A\).

The key to solving these types of problems usually involves determining \(k\). Depending on what your instructor wants, you can usually just start with the equation \(y = Ae^{kt}\), if you know that you have an exponential decay or growth problem.

Here is a quick video explaining this and showing a graph to give you a feel for these equations.

Half-life is the time it takes for half the substance to decay. The idea is to take the equation \(y = Ae^{kt}\), set the left side to \(A/2\) and solve for t. Notice that you don't have to know the initial amount A since in the equation \(A/2 = Ae^{kt}\), the A cancels leaving \(1/2 = e^{kt}\). You can then use basic logarithms to solve for t.

So it may not have occurred to you but have you thought that maybe it is possible there is another solution to the differential equation \(x'=ax\)? This question is part of a very deep discussion in differential equations involving existence and uniqueness. Here is a really good video that shows that the solution discussed on this page is unique.