6.6 Exponential and logarithmic equations  (Page 3/8)

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Solve $\text{\hspace{0.17em}}{2}^{x}={3}^{x+1}.$

$x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(2}{3}\right)}$

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 $\text{\hspace{0.17em}}t.$

1. Divide both sides of the equation by $\text{\hspace{0.17em}}A.$
2. Apply the natural logarithm of both sides of the equation.
3. Divide both sides of the equation by $\text{\hspace{0.17em}}k.$

Solve an equation of the form y = Ae kt

Solve $\text{\hspace{0.17em}}100=20{e}^{2t}.$

Solve $\text{\hspace{0.17em}}3{e}^{0.5t}=11.$

$t=2\mathrm{ln}\left(\frac{11}{3}\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}$

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 $\text{\hspace{0.17em}}2=-3{e}^{t}.$

Solving an equation that can be simplified to the form y = Ae kt

Solve $\text{\hspace{0.17em}}4{e}^{2x}+5=12.$

Solve $\text{\hspace{0.17em}}3+{e}^{2t}=7{e}^{2t}.$

$t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)$

Extraneous solutions

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.

Solving exponential functions in quadratic form

Solve $\text{\hspace{0.17em}}{e}^{2x}-{e}^{x}=56.$

Solve $\text{\hspace{0.17em}}{e}^{2x}={e}^{x}+2.$

$x=\mathrm{ln}2$

Does every logarithmic equation have a solution?

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 $\text{\hspace{0.17em}}x:$

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
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?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
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
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
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar