Is there any way to solve
$\text{\hspace{0.17em}}{2}^{x}={3}^{x}?$
Yes. The solution is
$0.$
Equations containing
e
One common type of exponential equations are those with base
$\text{\hspace{0.17em}}e.\text{\hspace{0.17em}}$ This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base
$\text{\hspace{0.17em}}e\text{\hspace{0.17em}}$ on either side, we can use the
natural logarithm to solve it.
Given an equation of the form
$\text{\hspace{0.17em}}y=A{e}^{kt}\text{,}$ solve for
$\text{\hspace{0.17em}}t.$
Divide both sides of the equation by
$\text{\hspace{0.17em}}A.$
Apply the natural logarithm of both sides of the equation.
Divide both sides of the equation by
$\text{\hspace{0.17em}}k.$
Does every equation of the form$\text{\hspace{0.17em}}y=A{e}^{kt}\text{\hspace{0.17em}}$have a solution?
No. There is a solution when
$\text{\hspace{0.17em}}k\ne 0,$ and when
$\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is
$\text{\hspace{0.17em}}2=\mathrm{-3}{e}^{t}.$
Solving an equation that can be simplified to the form
y =
Ae^{
kt }
Sometimes the methods used to solve an equation introduce an
extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.
No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.
Using the definition of a logarithm to solve logarithmic equations
We have already seen that every
logarithmic equation$\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(x\right)=y\text{\hspace{0.17em}}$ is equivalent to the exponential equation
$\text{\hspace{0.17em}}{b}^{y}=x.\text{\hspace{0.17em}}$ We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
For example, consider the equation
$\text{\hspace{0.17em}}{\mathrm{log}}_{2}\left(2\right)+{\mathrm{log}}_{2}\left(3x-5\right)=3.\text{\hspace{0.17em}}$ To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for
$\text{\hspace{0.17em}}x:$
Questions & Answers
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Period =2π
if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
For Plan A to reach $27/month to surpass Plan B's $26.50 monthly payment, you'll need 3,000 texts which will cost an additional $10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic.
Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation
of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15)
it's standard equation is x^2 + y^2/16 =1
tell my why is it only x^2? why is there no a^2?