How do you write an exponential equation that passes through \[\left( {0, - 2} \right)\]and \[\left( {2, - 50} \right)\]?
Answer
576.3k+ views
Hint: An exponential function is defined by the formula \[f\left( x \right) = {a^x}\], where the input variable x occurs as an exponent, in which “x” is a variable and “a” is a constant which is called the base of the function and it should be greater than 0. To write an exponential equation with the given points we need to consider a function such that, substituting the value of the given points we obtain the exponential equation.
Complete step by step solution:
An exponential equation that passes through \[\left( {0, - 2} \right)\]and \[\left( {2, - 50} \right)\].
Let, us set the exponential equation as:
\[f\left( x \right) = a \cdot {e^{bx}}\] ……………………. 1
In which a and b are constants to be found, hence from the given two points that the function is passing we know that:
At point \[\left( {0, - 2} \right)\], we have:
\[f\left( 0 \right) = - 2\]
Now, with respect to equation 1, applying\[f\left( 0 \right) = - 2\] we get:
\[ \Rightarrow a \cdot {e^{b \cdot 0}} = - 2\]
Since, we know that \[{e^0} = 1\], we get:
\[ \Rightarrow a \cdot 1 = - 2\]
\[ \Rightarrow a = - 2\]
Now, at point \[\left( {2, - 50} \right)\], we have:
\[f\left( 2 \right) = - 50\]
Now, with respect to equation 1, applying\[f\left( 2 \right) = - 50\] we get:
\[ \Rightarrow - 2 \cdot {e^{b \cdot 2}} = - 50\]
\[ \Rightarrow {e^{2b}} = \dfrac{{ - 50}}{{ - 2}}\]
Simplifying the terms, we get:
\[ \Rightarrow {e^{2b}} = 25\]
Apply ln on both sides of we get:
\[\ln {e^{2b}} = \ln 25\]
We know that \[ \ln \left( {{e^x}} \right) = \log \left( {{e^x}} \right) = x\], hence we get:
\[ \Rightarrow 2b = \ln 25\]
Shift the terms, to evaluate the value as:
\[ \Rightarrow b = \dfrac{{\ln 25}}{2}\]
We, know that \[\ln 25 = 3.21888..\], hence substituting it we get:
\[ \Rightarrow b = \dfrac{{3.21888}}{2}\]
\[ \Rightarrow b = 1.60944\]
Hence, the constants are known now i.e., \[a = - 2\] and \[b = 1.60944\], so substitute these values in the function i.e., equation 1 as:
\[f\left( x \right) = a \cdot {e^{bx}}\]
\[ \Rightarrow f\left( x \right) = - 2 \cdot {e^{1.60944 \cdot x}}\].
Note: The most commonly used exponential function base is e, which is approximately equal to 2.71828. The exponential curve depends on the exponential function \[f\left( x \right) = {a^x}\]and it depends on the value of the x, where \[a > 0\] and a is not equal to one and x is any real number. Exponential function having base 10 is known as a common exponential function.
Complete step by step solution:
An exponential equation that passes through \[\left( {0, - 2} \right)\]and \[\left( {2, - 50} \right)\].
Let, us set the exponential equation as:
\[f\left( x \right) = a \cdot {e^{bx}}\] ……………………. 1
In which a and b are constants to be found, hence from the given two points that the function is passing we know that:
At point \[\left( {0, - 2} \right)\], we have:
\[f\left( 0 \right) = - 2\]
Now, with respect to equation 1, applying\[f\left( 0 \right) = - 2\] we get:
\[ \Rightarrow a \cdot {e^{b \cdot 0}} = - 2\]
Since, we know that \[{e^0} = 1\], we get:
\[ \Rightarrow a \cdot 1 = - 2\]
\[ \Rightarrow a = - 2\]
Now, at point \[\left( {2, - 50} \right)\], we have:
\[f\left( 2 \right) = - 50\]
Now, with respect to equation 1, applying\[f\left( 2 \right) = - 50\] we get:
\[ \Rightarrow - 2 \cdot {e^{b \cdot 2}} = - 50\]
\[ \Rightarrow {e^{2b}} = \dfrac{{ - 50}}{{ - 2}}\]
Simplifying the terms, we get:
\[ \Rightarrow {e^{2b}} = 25\]
Apply ln on both sides of we get:
\[\ln {e^{2b}} = \ln 25\]
We know that \[ \ln \left( {{e^x}} \right) = \log \left( {{e^x}} \right) = x\], hence we get:
\[ \Rightarrow 2b = \ln 25\]
Shift the terms, to evaluate the value as:
\[ \Rightarrow b = \dfrac{{\ln 25}}{2}\]
We, know that \[\ln 25 = 3.21888..\], hence substituting it we get:
\[ \Rightarrow b = \dfrac{{3.21888}}{2}\]
\[ \Rightarrow b = 1.60944\]
Hence, the constants are known now i.e., \[a = - 2\] and \[b = 1.60944\], so substitute these values in the function i.e., equation 1 as:
\[f\left( x \right) = a \cdot {e^{bx}}\]
\[ \Rightarrow f\left( x \right) = - 2 \cdot {e^{1.60944 \cdot x}}\].
Note: The most commonly used exponential function base is e, which is approximately equal to 2.71828. The exponential curve depends on the exponential function \[f\left( x \right) = {a^x}\]and it depends on the value of the x, where \[a > 0\] and a is not equal to one and x is any real number. Exponential function having base 10 is known as a common exponential function.
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