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
$\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
Rewrite
$\text{\hspace{0.17em}}y=a{b}^{x}\text{\hspace{0.17em}}$ as
$\text{\hspace{0.17em}}y=a{e}^{\mathrm{ln}\left({b}^{x}\right)}.$
Use the power rule of logarithms to rewrite y as
$\text{\hspace{0.17em}}y=a{e}^{x\mathrm{ln}\left(b\right)}=a{e}^{\mathrm{ln}\left(b\right)x}.$
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
$\text{\hspace{0.17em}}y=2.5{(3.1)}^{x}\text{\hspace{0.17em}}$ so that this same function is written in the form
$\text{\hspace{0.17em}}y={A}_{0}{e}^{kx}.$
Change the function
$\text{\hspace{0.17em}}y=3{(0.5)}^{x}\text{\hspace{0.17em}}$ to one having
$\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ as the base.
If
$\text{}A={A}_{0}{e}^{kt},$$k<0,$ the half-life is
$\text{}t=-\frac{\mathrm{ln}(2)}{k}.$
Carbon-14 dating
$t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\right)}{-0.000121}.$ ${A}_{0}\text{}$$A\text{}$ is the amount of carbon-14 when the plant or animal died
$t\text{}$ is the amount of carbon-14 remaining today
is the age of the fossil in years
Doubling time formula
If
$\text{}A={A}_{0}{e}^{kt},$$k>0,$ the doubling time is
$\text{}t=\frac{\mathrm{ln}2}{k}$
Newton’s Law of Cooling
$T(t)=A{e}^{kt}+{T}_{s},$ where
$\text{}{T}_{s}\text{}$ is the ambient temperature,
$\text{}A=T(0)-{T}_{s},$ and
$\text{}k\text{}$ is the continuous rate of cooling.
Key concepts
The basic exponential function is
$\text{\hspace{0.17em}}f(x)=a{b}^{x}.\text{\hspace{0.17em}}$ If
$\text{\hspace{0.17em}}b>1,$ we have exponential growth; if
$\text{\hspace{0.17em}}0<b<1,$ 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] .
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?
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
Adu
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