Which of the following exponential equations could be represented by the table below?
| x | 3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| --- | ------ | ------ | ------ | ------ | ------ | ------ | --- | --- | ---- |
| y | 2.0014 | 2.0041 | 2.0123 | 2.037 | 2.111 | 2.333 | 3 | 5 | 11 |
A. [tex]y=\left(\frac{1}{3}\right)^{x-3}-3[/tex]
B. [tex]y=3^{x-2}-3[/tex]
C. [tex]y=3^{x-3}+2[/tex]
D. [tex]y=\left(\frac{1}{3}\right)^{x-3}-2[/tex]
Answer (1)
Substitute x values from the table into each of the four equations.
Compare the calculated y values with the y values in the table.
Observe that the equation y = 3 x − 3 + 2 matches the table values.
Conclude that the exponential equation represented by the table is y = 3 x − 3 + 2 .
Explanation
Problem Analysis We are given a table of x and y values and four exponential equations. Our goal is to determine which equation best represents the data in the table. We will substitute the x values from the table into each equation and compare the resulting y values with the values in the table.
Testing Equation 1 Let's test the first equation: y = ( 3 1 ) x − 3 − 3 . We will substitute x = 0 and x = 3 into this equation. For x = 0 : y = ( 3 1 ) 0 − 3 − 3 = ( 3 1 ) − 3 − 3 = 3 3 − 3 = 27 − 3 = 24 . This is significantly different from the table value of y = 2.037 when x = 0 .
For x = 3 : y = ( 3 1 ) 3 − 3 − 3 = ( 3 1 ) 0 − 3 = 1 − 3 = − 2 . This is different from the table value of y = 3 when x = 3 .
Testing Equation 2 Now, let's test the second equation: y = 3 x − 2 − 3 . We will substitute x = 0 and x = 3 into this equation. For x = 0 : y = 3 0 − 2 − 3 = 3 − 2 − 3 = 9 1 − 3 = 9 1 − 9 27 = − 9 26 ≈ − 2.89 . This is significantly different from the table value of y = 2.037 when x = 0 .
For x = 3 : y = 3 3 − 2 − 3 = 3 1 − 3 = 3 − 3 = 0 . This is different from the table value of y = 3 when x = 3 .
Testing Equation 3 Next, let's test the third equation: y = 3 x − 3 + 2 . We will substitute x = 0 , 1 , 2 , 3 , 4 , 5 into this equation and compare with the table values. For x = 0 : y = 3 0 − 3 + 2 = 3 − 3 + 2 = 27 1 + 2 = 27 1 + 27 54 = 27 55 ≈ 2.037 . This matches the table value. For x = 1 : y = 3 1 − 3 + 2 = 3 − 2 + 2 = 9 1 + 2 = 9 1 + 9 18 = 9 19 ≈ 2.111 . This matches the table value. For x = 2 : y = 3 2 − 3 + 2 = 3 − 1 + 2 = 3 1 + 2 = 3 1 + 3 6 = 3 7 ≈ 2.333 . This matches the table value. For x = 3 : y = 3 3 − 3 + 2 = 3 0 + 2 = 1 + 2 = 3 . This matches the table value. For x = 4 : y = 3 4 − 3 + 2 = 3 1 + 2 = 3 + 2 = 5 . This matches the table value. For x = 5 : y = 3 5 − 3 + 2 = 3 2 + 2 = 9 + 2 = 11 . This matches the table value.
Testing Equation 4 Finally, let's test the fourth equation: y = ( 3 1 ) x − 3 − 2 . We will substitute x = 0 and x = 3 into this equation. For x = 0 : y = ( 3 1 ) 0 − 3 − 2 = ( 3 1 ) − 3 − 2 = 3 3 − 2 = 27 − 2 = 25 . This is significantly different from the table value of y = 2.037 when x = 0 .
For x = 3 : y = ( 3 1 ) 3 − 3 − 2 = ( 3 1 ) 0 − 2 = 1 − 2 = − 1 . This is different from the table value of y = 3 when x = 3 .
Conclusion By substituting the x values from the table into each equation, we found that the third equation, y = 3 x − 3 + 2 , produces y values that match the table values. Therefore, the exponential equation represented by the table is y = 3 x − 3 + 2 .
Examples
Exponential equations are used to model various real-world phenomena, such as population growth, radioactive decay, and compound interest. For example, if a bacteria colony starts with 100 bacteria and doubles every hour, the population can be modeled by the equation P ( t ) = 100 × 2 t , where P ( t ) is the population after t hours. Understanding exponential equations allows us to predict future values based on current trends.
