For the following exercises, find the formula for an exponential function that passes through th

For the following exercises, find the formula for an...

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Key Concepts

Logarithmic Techniques

Logarithmic techniques allow you to isolate and solve for the exponential term. When you have an equation where the variable is in the exponent, taking the logarithm of both sides enables you to bring the exponent down, thereby simplifying the equation and making it possible to solve for the unknown coefficient.

Systems of Equations

In this context, a system of equations is developed by inserting the x and y values from each point into the exponential equation. The resulting system is solved through various algebraic methods, such as substitution or elimination, to find the unknown parameters.

Exponential Functions

Exponential functions are equations in which the variable appears in the exponent, typically expressed in the form f(x) = ab^x. This form is used to model growth or decay processes, with 'a' representing the initial value and 'b' denoting the constant multiplicative rate of change.

Finding Parameters Using Two Points

Determining the parameters of an exponential function using two points involves substituting the given coordinates into the general formula to create a system of equations. By solving this system, one can uniquely identify the values of the initial value 'a' and the base 'b', thus obtaining the specific exponential function that fits the data.

Transcript

00:01 So we want to find the exponential equation represented by f of x is equal to a times b to the x.

00:08 That goes through the points negative 1, 3 over 2, and the point 324.

00:14 So to go ahead and start off, what we can do is we can use one of these points.

00:18 I'll use 324 to find out how a and b relate to each other.

00:23 So to do this, we can set 24 equal to a times b to the third.

00:33 24 is equal to a times b to the third.

00:38 Now, if we wanted to isolate a in this situation, we could do that pretty easily by dividing both sides by b to the third.

00:46 So we would get 24 over b to the third is equal to a.

00:52 24 over b to the third is equal to a.

00:58 And we can use this in combination with another equation to fully solve for b, and then we can solve for a.

01:09 So what we'll want to do next is we're going to want to use this point now, negative 1, 3 over 2.

01:15 And so we're going to want to do 3 over 2.

01:18 3 over 2 is equal to is equal to a, which we're going to 1, which we're going to 1 ,000.

01:24 We know is 24 over b to the 3rd, so 24 over b to the 3rd times b to the negative 1.

01:38 And we can fully simplify this out.

01:40 So the first thing that we'll notice is that 24 over b to the 3 times b to the negative 1 is just 24 over b to the 4th...

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