
How do you find the exponential function formula for the points $h(0) = 3$ and $h(1) = 15$ ?
Answer
556.8k+ views
Hint: The values of $x$ and $y$ are given in the form $h(x) = y$ . At first, we will use the formula $h(x) = a{b^x}$
where $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ . Then we will find the values of $a$ and $b$ . Then put these values in the formula.
Formula used: $h(x) = a{b^x}$; $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ .
Complete step-by-step solution:
We have;
$h(0) = 3$ …… $(1)$
And $h(1) = 15$ ……. $(2)$
These two values are in the form $h(x) = y$ .
Hence, $h(0) = 3$ means $y$ gives the value $3$ when $x$ is equal to $0$ .
Similarly, $h(1) = 15$ means $y$ gives the value $15$ when $x$ is equal to $1$ .
Now we use the formula $h(x) = a{b^x}$ ; $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ and find the values of $a$ and $b$ .
Comparing $h(x) = a{b^x}$ with $(1)$ we will get;
$3 = a{b^0}$
As we know that ${b^0} = 1$ ;
$\Rightarrow 3 = a \cdot 1$
$\Rightarrow a = 3$
Comparing $h(x) = a{b^x}$ with $(2)$ we will get;
$15 = a{b^1}$
As we know that ${b^1} = b$ ;
$\Rightarrow 15 = a \cdot b$
As we have seen $a = 3$ ;
$\Rightarrow 15 = 3 \cdot b$
Dividing both sides with $3$ we will get;
$\Rightarrow b = \dfrac{{15}}{3}$
$\Rightarrow b = 5$
Now we will put the values of $a$ and $b$ in the formula and we will get;
$h(x) = 3 \cdot {5^x}$
$h(x) = 3 \cdot {5^x}$ is the required formula of the exponential function where $x$ belongs to the set of real numbers.
Note: The actual well known exponential function is $h(x) = {e^x}$ ; where $2 < e < 3$ . But $e$ has a fixed value.
In this kind of problem there will always be two points given in the two-dimensional coordinate system. So that we can get the actual function. Students should always remember to put the values of constants in the formula and keep the variable the same as it is given to us.
where $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ . Then we will find the values of $a$ and $b$ . Then put these values in the formula.
Formula used: $h(x) = a{b^x}$; $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ .
Complete step-by-step solution:
We have;
$h(0) = 3$ …… $(1)$
And $h(1) = 15$ ……. $(2)$
These two values are in the form $h(x) = y$ .
Hence, $h(0) = 3$ means $y$ gives the value $3$ when $x$ is equal to $0$ .
Similarly, $h(1) = 15$ means $y$ gives the value $15$ when $x$ is equal to $1$ .
Now we use the formula $h(x) = a{b^x}$ ; $a \in$ set of real numbers and $b \in$ set of real numbers except $0$ and find the values of $a$ and $b$ .
Comparing $h(x) = a{b^x}$ with $(1)$ we will get;
$3 = a{b^0}$
As we know that ${b^0} = 1$ ;
$\Rightarrow 3 = a \cdot 1$
$\Rightarrow a = 3$
Comparing $h(x) = a{b^x}$ with $(2)$ we will get;
$15 = a{b^1}$
As we know that ${b^1} = b$ ;
$\Rightarrow 15 = a \cdot b$
As we have seen $a = 3$ ;
$\Rightarrow 15 = 3 \cdot b$
Dividing both sides with $3$ we will get;
$\Rightarrow b = \dfrac{{15}}{3}$
$\Rightarrow b = 5$
Now we will put the values of $a$ and $b$ in the formula and we will get;
$h(x) = 3 \cdot {5^x}$
$h(x) = 3 \cdot {5^x}$ is the required formula of the exponential function where $x$ belongs to the set of real numbers.
Note: The actual well known exponential function is $h(x) = {e^x}$ ; where $2 < e < 3$ . But $e$ has a fixed value.
In this kind of problem there will always be two points given in the two-dimensional coordinate system. So that we can get the actual function. Students should always remember to put the values of constants in the formula and keep the variable the same as it is given to us.
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