# 6.7 Exponential and logarithmic models  (Page 7/16)

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Does a linear, exponential, or logarithmic model best fit the data in [link] ? Find the model.

 $x$ 1 2 3 4 5 6 7 8 9 $y$ 3.297 5.437 8.963 14.778 24.365 40.172 66.231 109.196 180.034

Exponential. $\text{\hspace{0.17em}}y=2{e}^{0.5x}.$

## Expressing an exponential model in base e

While powers and logarithms of any base can be used in modeling, the two most common bases are $\text{\hspace{0.17em}}10\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}e.\text{\hspace{0.17em}}$ In science and mathematics, the base $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ is often preferred. We can use laws of exponents and laws of logarithms to change any base to base $\text{\hspace{0.17em}}e.$

Given a model with the form $\text{\hspace{0.17em}}y=a{b}^{x},$ change it to the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$

1. Rewrite $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}.$
2. Use the power rule of logarithms to rewrite y as $\text{\hspace{0.17em}}y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}.$
3. Note that $\text{\hspace{0.17em}}a={A}_{0}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}k=\mathrm{ln}\left(b\right)\text{\hspace{0.17em}}$ in the equation $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$

## Changing to base e

Change the function $\text{\hspace{0.17em}}y=2.5{\left(3.1\right)}^{x}\text{\hspace{0.17em}}$ so that this same function is written in the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$

The formula is derived as follows

Change the function $\text{\hspace{0.17em}}y=3{\left(0.5\right)}^{x}\text{\hspace{0.17em}}$ to one having $\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the base.

$y=3{e}^{\left(\mathrm{ln}0.5\right)x}$

Access these online resources for additional instruction and practice with exponential and logarithmic models.

## Key equations

 Half-life formula If $k<0,$ the half-life is Carbon-14 dating $t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}.$ is the amount of carbon-14 when the plant or animal died is the amount of carbon-14 remaining today is the age of the fossil in years Doubling time formula If $k>0,$ the doubling time is Newton’s Law of Cooling $T\left(t\right)=A{e}^{kt}+{T}_{s},$ where is the ambient temperature, and is the continuous rate of cooling.

## Key concepts

• The basic exponential function is $\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}b>1,$ we have exponential growth; if $\text{\hspace{0.17em}}0 we have exponential decay.
• We can also write this formula in terms of continuous growth as $\text{\hspace{0.17em}}A={A}_{0}{e}^{kx},$ where $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is the starting value. If $\text{\hspace{0.17em}}{A}_{0}\text{\hspace{0.17em}}$ is positive, then we have exponential growth when $\text{\hspace{0.17em}}k>0\text{\hspace{0.17em}}$ and exponential decay when $\text{\hspace{0.17em}}k<0.\text{\hspace{0.17em}}$ See [link] .
• In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay. See [link] .
• We can find the age, $\text{\hspace{0.17em}}t,$ of an organic artifact by measuring the amount, $\text{\hspace{0.17em}}k,$ of carbon-14 remaining in the artifact and using the formula $\text{\hspace{0.17em}}t=\frac{\mathrm{ln}\left(k\right)}{-0.000121}\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ See [link] .
• Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay. See [link] .
• We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time. See [link] .
• We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors. See [link] .
• We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data. See [link] .
• Any exponential function with the form $\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ can be rewritten as an equivalent exponential function with the form $\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k=\mathrm{ln}b.\text{\hspace{0.17em}}$ See [link] .

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
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Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
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I think you should say "28 terms" instead of "28th term"
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the 28th term is 175
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