Question

Write the exponential function. please show work

Answer 1

`y=10e^(x\ln4)=10(4)^x`

`y=0.4e^(-x\ln3)=0.4(3)^(-x)`

Explanation:

`\boxed{\begin{minipage}{9 cm}\underline{General form of an Exponential Function with base $e$}\\\\$y=ae^(kx)$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the initial value ($y$-intercept). \\ \phantom{ww}$\bullet$ $e$ is Euler's number. \\ \phantom{ww}$\bullet$ $k$ is some constant.\\\end{minipage}}`

Given points:

• (1, 40)
• (3, 640)

Substitute the points into the formula to create two equations:

`\implies ae^(k)=40`

`\implies ae^(3k)=640`

Divide the equations to eliminate a and solve for k:

`\implies (ae^(3k))/(ae^(k))=(640)/(40)`

`\implies (e^(3k))/(e^(k))=16`

`\implies e^(2k)=16`

`\implies \ln e^(2k)=\ln 16`

`\implies \ln e^(2k)=2\ln 4`

`\implies 2k=2\ln 4`

`\implies k=\ln 4`

Substitute the found value of k into the formula, along with one of the points, and solve for a:

`\implies ae^(\ln 4)=40`

`\implies 4a=40`

`\implies a=10`

Therefore, the exponential function for points (1, 40) and (3, 640) is:

`\implies y=10e^(x\ln4)`

`\implies y=10(e^(\ln 4))^x`

`\implies y=10(4)^x`

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Given points from the graph:

• (-3, 10.8)
• (-2, 3.6)

Substitute the points into the formula to create two equations:

`\implies ae^(-2k)=3.6`

`\implies ae^(-3k)=10.8`

Divide the equations to eliminate a and solve for k:

`\implies (ae^(-3k))/(ae^(-2k))=(10.8)/(3.6)`

`\implies (e^(-3k))/(e^(-2k))=3`

`\implies e^(-k)=3`

`\implies \ln e^(-k)=\ln 3`

`\implies -k=\ln 3`

`\implies k=-\ln 3`

Substitute the found value of k into the formula, along with one of the points, and solve for a:

`\implies ae^(2\ln 3)=3.6`

`\implies 9a=3.6`

`\implies a=0.4`

Therefore, the exponential function for points (1, 40) and (3, 640) is:

`\implies y=0.4e^(-x\ln3)`

`\implies y=0.4(e^(\ln 3))^(-x)`

`\implies y=0.4(3)^(-x)`