# 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] .

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar