Question

the scatterplot shows the relationship between two variables, x and y. an equation for the exponential model shown can be written as y = a(b)^x, where a and b are positive constants. which of the following is closest to the value of b? a. 0.84 b. 1.84 c. 1.74 d. 1.94

Explanation:

Step1: Select two points on the curve

Let's take two points \((x_1,y_1)=(20,2)\) and \((x_2,y_2)=(40,4)\) from the scatter - plot. The exponential equation is \(y = a(b)^x\).
For the point \((20,2)\): \(2=a(b)^{20}\). For the point \((40,4)\): \(4=a(b)^{40}\).

Step2: Divide the second equation by the first equation

\(\frac{4}{2}=\frac{a(b)^{40}}{a(b)^{20}}\).
Simplify the left - hand side to get \(2\), and use the exponent rule \(\frac{b^m}{b^n}=b^{m - n}\) on the right - hand side. So \(2 = b^{40 - 20}=b^{20}\).

Step3: Solve for \(b\)

Take the 20th root of both sides. \(b = 2^{\frac{1}{20}}\).
Using a calculator, \(b=2^{\frac{1}{20}}\approx1.035\). Another way is to use the general form. If we assume the initial point \((x = 0,y=a)\) (not shown on the graph but from the equation \(y=a(b)^x\)). Let's use two non - zero \(x\) values.
Let's rewrite the equation \(y=a(b)^x\) as \(\ln y=\ln a+x\ln b\). If we take two points \((x_1,y_1)\) and \((x_2,y_2)\): \(\ln y_1=\ln a+x_1\ln b\) and \(\ln y_2=\ln a+x_2\ln b\).
Subtract the first equation from the second: \(\ln y_2-\ln y_1=(x_2 - x_1)\ln b\), so \(\ln b=\frac{\ln y_2-\ln y_1}{x_2 - x_1}\).
Using points \((20,2)\) and \((40,4)\): \(\ln b=\frac{\ln4-\ln2}{40 - 20}=\frac{\ln\frac{4}{2}}{20}=\frac{\ln2}{20}\). Then \(b = e^{\frac{\ln2}{20}}\). Since \(e^{\ln2}=2\), \(b = 2^{\frac{1}{20}}\approx1.035\).
If we assume the points \((x = 20,y = 2)\) and \((x=60,y = 8)\)
\(\ln b=\frac{\ln8-\ln2}{60 - 20}=\frac{\ln\frac{8}{2}}{40}=\frac{\ln4}{40}=\frac{2\ln2}{40}=\frac{\ln2}{20}\)
If we use the fact that when \(x\) increases by 20, \(y\) doubles approximately.
Let \(x\) change from \(x_1\) to \(x_1 + 20\), \(y_2=a(b)^{x_1+20}\) and \(y_1=a(b)^{x_1}\), \(\frac{y_2}{y_1}=b^{20}\). If \(\frac{y_2}{y_1}\approx2\), \(b = 2^{\frac{1}{20}}\approx1.035\) which is closest to \(1.084\) among the given options.

Answer:

a. \(0.84\)