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Chapter 3: Problem 28

Find the exponential model that fits the points shown in the graph or table. $$\begin{array}{|c|c|c|} \hline x & 0 & 3 \\ \hline y & 1 & \frac{1}{4} \\ \hline \end{array}$$

Short Answer

Expert verified
The exponential model that fits the points given in the table is \(y = (0.629)^x\).

Step by step solution

01

Formulating the system of equations

Let's formulate the system of equations from the given points by plugging them into the general formula for exponential functions \(y = ab^x\). The two equations will be as follows: \(a \cdot b^0 = 1\) from the point (0,1) and \(a \cdot b^3 = 1/4\) from the point (3,1/4).
02

Solving the first equation

We notice that any number raised to power 0 equals 1, so the first equation simplifies to \(a=1\). Then our function model takes the form \(y=b^x\).
03

Substituting \(a\) into the second equation

Now let's substitute \(a=1\) into the second equation to find \(b\). The second equation becomes \(b^3 = 1/4\). Now, we need to solve this equation for \(b\) by taking the cubic root of both sides.
04

Solving for \(b\)

To solve the equation \(b^3 = 1/4\) for \(b\), we take the cubic root of both sides to get \(b = \sqrt[3]{1/4}\). Using a calculator, we find \(b = \sqrt[3]{1/4} \approx 0.629\).
05

Construct the final function

Finally, we substitute the found values of \(a\) and \(b\) into the general form. This gives us the required exponential model: \(y = (0.629)^x\).

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Most popular questions from this chapter

In Exercises \(97-102,\) determine whether the statement is true or false given that \(f(x)=\ln x .\) Justify your answer. $$\sqrt{f(x)}=\frac{1}{2} f(x)$$

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Is it possible for a logarithmic equation to have more than one extraneous solution? Explain.

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The graph of \(f(x)=\log _{3} x\) contains the point \((27,3)\)
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