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Sagot :
To determine which set of ordered pairs could be generated by an exponential function, we need to analyze each set to see if they fit the general form of an exponential function [tex]\( y = a \cdot b^x \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are constants.
Set 1: [tex]\((1, 1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
Let's see if we can derive a consistent set of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] that match these points:
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex] to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute [tex]\( a \)[/tex] into second equation:
[tex]\[ \frac{1}{2} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{2} = b \][/tex]
Substituting [tex]\( b = \frac{1}{2} \)[/tex] in [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = 1 \][/tex]
[tex]\[ a = 2 \][/tex]
Now let's check the other points:
For point [tex]\((3, \frac{1}{3})\)[/tex]:
[tex]\[ \frac{1}{3} = 2 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{3} = \frac{2}{8} = \frac{1}{4} \][/tex]
This contradicts, so Set 1 is not exponential.
Set 2: [tex]\((1, 1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = b \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{4} = 1 \][/tex]
[tex]\[ a = 4 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{9})\)[/tex]:
[tex]\[ \frac{1}{9} = 4 \cdot \left(\frac{1}{4}\right)^3 \][/tex]
[tex]\[ \frac{1}{9} = 4 \cdot \frac{1}{64} = \frac{1}{16} \][/tex]
This contradicts, so Set 2 is not exponential.
Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = \frac{1}{2} \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{8})\)[/tex]:
[tex]\[ \frac{1}{8} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{8} = \frac{1}{8} \][/tex]
For point [tex]\((4, \frac{1}{16})\)[/tex]:
[tex]\[ \frac{1}{16} = 1 \cdot \left(\frac{1}{2}\right)^4 \][/tex]
[tex]\[ \frac{1}{16} = \frac{1}{16} \][/tex]
All points fit, so Set 3 is generated by an exponential function.
Set 4: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{6}\right), \left(4, \frac{1}{8}\right)\)[/tex]
Using similar steps:
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{6})\)[/tex]:
[tex]\[ \frac{1}{6} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{6} \neq \frac{1}{8} \][/tex]
This contradicts, so Set 4 is not exponential.
Thus, the correct set of ordered pairs that could be generated by an exponential function is Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex].
Set 1: [tex]\((1, 1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
Let's see if we can derive a consistent set of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] that match these points:
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex] to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute [tex]\( a \)[/tex] into second equation:
[tex]\[ \frac{1}{2} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{2} = b \][/tex]
Substituting [tex]\( b = \frac{1}{2} \)[/tex] in [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = 1 \][/tex]
[tex]\[ a = 2 \][/tex]
Now let's check the other points:
For point [tex]\((3, \frac{1}{3})\)[/tex]:
[tex]\[ \frac{1}{3} = 2 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{3} = \frac{2}{8} = \frac{1}{4} \][/tex]
This contradicts, so Set 1 is not exponential.
Set 2: [tex]\((1, 1), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{9}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, 1)\)[/tex]:
[tex]\[ 1 = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = 1 \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = 1 \)[/tex]:
[tex]\[ a = \frac{1}{b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = b \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{4} = 1 \][/tex]
[tex]\[ a = 4 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{9})\)[/tex]:
[tex]\[ \frac{1}{9} = 4 \cdot \left(\frac{1}{4}\right)^3 \][/tex]
[tex]\[ \frac{1}{9} = 4 \cdot \frac{1}{64} = \frac{1}{16} \][/tex]
This contradicts, so Set 2 is not exponential.
Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex]
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b^1 \][/tex]
[tex]\[ a \cdot b = \frac{1}{2} \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{8})\)[/tex]:
[tex]\[ \frac{1}{8} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{8} = \frac{1}{8} \][/tex]
For point [tex]\((4, \frac{1}{16})\)[/tex]:
[tex]\[ \frac{1}{16} = 1 \cdot \left(\frac{1}{2}\right)^4 \][/tex]
[tex]\[ \frac{1}{16} = \frac{1}{16} \][/tex]
All points fit, so Set 3 is generated by an exponential function.
Set 4: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{6}\right), \left(4, \frac{1}{8}\right)\)[/tex]
Using similar steps:
For point [tex]\((1, \frac{1}{2})\)[/tex]:
[tex]\[ \frac{1}{2} = a \cdot b \][/tex]
For point [tex]\((2, \frac{1}{4})\)[/tex]:
[tex]\[ \frac{1}{4} = a \cdot b^2 \][/tex]
Using [tex]\( a \cdot b = \frac{1}{2} \)[/tex]:
[tex]\[ a = \frac{1}{2b} \][/tex]
Substitute:
[tex]\[ \frac{1}{4} = \frac{1}{2b} \cdot b^2 \][/tex]
[tex]\[ \frac{1}{4} = \frac{b}{2} \][/tex]
[tex]\[ b = \frac{1}{2} \][/tex]
Now [tex]\( a \)[/tex]:
[tex]\[ a \cdot \frac{1}{2} = \frac{1}{2} \][/tex]
[tex]\[ a = 1 \][/tex]
Check other points:
For point [tex]\((3, \frac{1}{6})\)[/tex]:
[tex]\[ \frac{1}{6} = 1 \cdot \left(\frac{1}{2}\right)^3 \][/tex]
[tex]\[ \frac{1}{6} \neq \frac{1}{8} \][/tex]
This contradicts, so Set 4 is not exponential.
Thus, the correct set of ordered pairs that could be generated by an exponential function is Set 3: [tex]\(\left(1, \frac{1}{2}\right), \left(2, \frac{1}{4}\right), \left(3, \frac{1}{8}\right), \left(4, \frac{1}{16}\right)\)[/tex].
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