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
can you not take the square root of a negative number
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