Model the data using an exponential function f(x)=Ab^x begin{array}{|l|l|l|}hline x&0&1&2hline f(x)&125&25&5hlineend{array}

Model the data using an exponential function f(x)=Ab^x begin{array}{|l|l|l|}hline x&0&1&2hline f(x)&125&25&5hlineend{array}

Question
Exponential models
asked 2021-03-07
Model the data using an exponential function \(f(x)=Ab^x\)
\(\begin{array}{|l|l|l|}\hline x&0&1&2\\\hline f(x)&125&25&5\\\hline\end{array}\)

Answers (1)

2021-03-08

Given- Consider the data,
\(\begin{array}{|l|l|l|}\hline x&0&1&2\\\hline f(x)&125&25&5\\\hline\end{array}\)
To model- The above data using an exponential function \(f(x)=Ab^x\)
Explanation- As per the question, we have to model the above data using the given exponential expression as,
\(f(x)=Ab^x\) (1)
Now, using first value \(x=0\) and \(f(x)=125\), we get from the above expression,
\(f(0)=Ab^0\)
\(125=A\cdot1\)
\(125=A\)
So, equation (1) becomes,
\(f(x)=125b^x\)
Now, using second value \(x=1\) and \(f(x)=25\), we get from the above expression,
\(f(1)=125b^1\) (2)
\(25=125\cdot b\)
The above expression can be written as,
\(b=\frac{25}{125}\)
\(b=\frac{1}{5}\)
So, equation (2), becomes,
\(f(x)=125(\frac{1}{5})^x\) (3)
Now, to check the above expression, we can use the third value as \(x=2\) and \(f(x)=5\), we get,
\(5=125\cdot(\frac{1}{5})^2\)
\(5=125\cdot\frac{1}{25}\)
\(5=5\)
So, the expression \(f(x)=125(\frac{1}{5})^x\) is correct expression.
Answer- Hence, the exponential expression which models the given data is
\(f(x)=125(\frac{1}{5})^x\)

0

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