Find the exponential model that fits the points shown in the graph. (Round the exponent to four decimal places)

Question
Exponents and radicals
asked 2020-11-07
Find the exponential model that fits the points shown in the graph. (Round the exponent to four decimal places)

Answers (1)

2020-11-08
Given points on the exponential model: (0,1) and (5,16).
General equation of an exponential model: y=aebx
Evaluate the general equation at x=0 and x=5:
\(\displaystyle{y}={a}{e}^{{b}}{\left({0}\right)}={a}{e}^{{0}}={a}{\left({1}\right)}={a}\)
\(\displaystyle{y}={a}{e}^{{b}}{\left({5}\right)}={a}{e}^{{5}}{b}\)
We require y=1 when x=0 and y=16 when x=5 (as the points (0,1) and (5,16) need to lie on the exponential model).
a=1
\(\displaystyle{a}{e}^{{5}}{b}={16}\)
By the first equation, we thus know that a=1. Let us replace \(\displaystyle{a}{e}^{{5}}{b}={16}\) by 1 in the second equation:
\(\displaystyle{e}^{{5}}{b}={16}\)
Take the natural logarithm from each side of the previous equation (also using that the natural logarithm is the inverse of the exponential).
\(\displaystyle{5}{b}={\ln{{16}}}\)
Let us use the power property of logarithms (lna^b=blna) along with \(\displaystyle{16}={2}^{{4}}:\)
\(\displaystyle{5}{b}={4}{\ln{{2}}}\)
Divide each side of the previous equation by 5:
\(\displaystyle{b}=\frac{{{4}{\ln{{2}}}}}{{5}}\)
Replacing a by 1 and b by \(\displaystyle{4}\frac{{\ln{{2}}}}{{5}}\) in the general equation of the exponential model, we then obtain:
\(\displaystyle{y}={e}^{{{4}\frac{{\ln{{2}}}}{{5}}}}{x}\)
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