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] .
Questions & Answers
can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
Someone should please solve it for me
Add 2over ×+3 +y-4 over 5
simplify (×+a)with square root of two -×root 2 all over a
multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15
Second one, I got Root 2
Third one, I got 1/(y to the fourth power)
I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
graph the following linear equation using intercepts method.
2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b
you were already given the 'm' and 'b'.
so..
y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line.
where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2
2=3x
x=3/2
then .
y=3/2X-2
I think
Given
co ordinates for x
x=0,(-2,0)
x=1,(1,1)
x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â