Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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 your visit. We are dedicated to helping you find the information you need, whenever you need it. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.