Is there any way to solve
$\text{\hspace{0.17em}}{2}^{x}={3}^{x}?$
Yes. The solution is
$\text{\hspace{0.17em}}x=0.$
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
$\text{\hspace{0.17em}}x:$
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