Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Ask your questions and receive precise answers from experienced professionals across different disciplines. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Sure, let's find the exponential function that passes through the points (5, 5) and (8, 135).
An exponential function can be written in the form:
[tex]\[ y = a \cdot b^x \][/tex]
To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we can use the given points (5, 5) and (8, 135).
First, rewrite the function for the two points:
1. For (5, 5):
[tex]\[ 5 = a \cdot b^5 \][/tex]
2. For (8, 135):
[tex]\[ 135 = a \cdot b^8 \][/tex]
Now, we have two equations:
[tex]\[ \begin{cases} 5 = a \cdot b^5 \\ 135 = a \cdot b^8 \end{cases} \][/tex]
To solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex], divide the second equation by the first to eliminate [tex]\(a\)[/tex]:
[tex]\[ \frac{135}{5} = \frac{a \cdot b^8}{a \cdot b^5} \][/tex]
Simplify the right-hand side:
[tex]\[ 27 = b^{8-5} \\ 27 = b^3 \\ \][/tex]
Now, solve for [tex]\(b\)[/tex]:
[tex]\[ b = \sqrt[3]{27} \\ b = 3 \][/tex]
With [tex]\(b\)[/tex] known, substitute [tex]\(b\)[/tex] back into one of the original equations to solve for [tex]\(a\)[/tex]:
Using [tex]\(5 = a \cdot 3^5\)[/tex]:
[tex]\[ 5 = a \cdot 243 \\ a = \frac{5}{243} \\ a \approx 0.0205761 \][/tex]
So, the values are:
[tex]\[ a \approx 0.0205761 \\ b = 3 \][/tex]
Thus, the exponential function that passes through the points (5, 5) and (8, 135) is:
[tex]\[ y = 0.0205761 \cdot 3^x \][/tex]
An exponential function can be written in the form:
[tex]\[ y = a \cdot b^x \][/tex]
To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex], we can use the given points (5, 5) and (8, 135).
First, rewrite the function for the two points:
1. For (5, 5):
[tex]\[ 5 = a \cdot b^5 \][/tex]
2. For (8, 135):
[tex]\[ 135 = a \cdot b^8 \][/tex]
Now, we have two equations:
[tex]\[ \begin{cases} 5 = a \cdot b^5 \\ 135 = a \cdot b^8 \end{cases} \][/tex]
To solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex], divide the second equation by the first to eliminate [tex]\(a\)[/tex]:
[tex]\[ \frac{135}{5} = \frac{a \cdot b^8}{a \cdot b^5} \][/tex]
Simplify the right-hand side:
[tex]\[ 27 = b^{8-5} \\ 27 = b^3 \\ \][/tex]
Now, solve for [tex]\(b\)[/tex]:
[tex]\[ b = \sqrt[3]{27} \\ b = 3 \][/tex]
With [tex]\(b\)[/tex] known, substitute [tex]\(b\)[/tex] back into one of the original equations to solve for [tex]\(a\)[/tex]:
Using [tex]\(5 = a \cdot 3^5\)[/tex]:
[tex]\[ 5 = a \cdot 243 \\ a = \frac{5}{243} \\ a \approx 0.0205761 \][/tex]
So, the values are:
[tex]\[ a \approx 0.0205761 \\ b = 3 \][/tex]
Thus, the exponential function that passes through the points (5, 5) and (8, 135) is:
[tex]\[ y = 0.0205761 \cdot 3^x \][/tex]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.