If you wanted to include imaginary solutions, you could write -3 in its polar form (3e^ipi) and then take the natural log of that to get another solution, ln3+ipi
Essentially, it's just a hidden quadratic equation. Let capital E = e^x. This means e^(-x) is the same thing as 1/E. The given equation, e^x - 12*e^(-x) - 1 = 0, can therefore be rewritten as: E - 12/E - 1 = 0 Multiply thru by E, to clear the fraction: E^2 - E - 12 = 0 And solve the quadratic for E. E = 1/2 +/- sqrt(1/4 - (-12)) E = 4, and E = -3 are the two solutions We only want the positive solution, since e^x is never negative for real values of x. This means, e^x = 4, so we use natural log to invert the function and find x = ln(4).
Nice problem. On the last step, you can reexpress ln(4) as 2ln(2) and since ln(2) ~= 0.693 is more familiar to a student and more likely memorized than ln(4), we would not need a calculator.
I am watching many math solving clips on TH-cam, but as a student my concern was more with; how to I derive such an equation from observing my surrounding or something. That skill is more rewarding than solving it, in my opinion, because there are so many other tools to solve math problems right now. Take this equation for instance. Assuming that an entity has this behavior, how do I observe that and translate this behavior into this/a mathematical format. That's what I have always been looking for in math class but was never taught that in school.
That is a good question. The short answer is to become familiar with typical applications of each of the parent function types, such as linear equations, quadratics, exponentials, and trig. A common application of exponential functions is growth or decay. It's common for situations to either grow or decay, in proportion to the amount you have at any given instant. Population grows exponentially (at least initially). We could give the number of kids each couple has (suppose 3), and an average age of parents at birth (suppose 30). Since 2 kids replace the parents, this means only one child per woman, contributes to population growth. For this data, every 30 years, the population increases by a factor of 3/2. We can write the equation P(t) = P0 *(3/2)^(t/30), where P0 is the initial population at t=0. You could also write this as P(t) = P0 * e^(t*ln(3/2)/30). Temperature decays exponentially. Consider cup of tea, initially at 90 Celsius, and you take it outside for your morning walk to work on a cold 0 C morning. After 1 minute, suppose it cools to 80 C. You are interested in its temperature (big T) as a function of time in minutes (little t). This allows us to construct the equation, T(t) = 90*e^(-k*t), where k is an unknown constant. Newton's law of cooling is the physical basis for this equation, which tells us that cooling rates are proportional to temperature difference. Exponential functions are the solution to this constraint. Construct another version, at t=1 min, and we can solve for k. Our function becomes T(t) = 90*e^(-ln(9/8)*t). Maybe you'd like to know when it will be 50C, so it's cold enough to drink; answer: t=5 min.
Not every example in a math textbook, necessarily comes from a real world example. Some come from just curiosity to explore how a function works. It could be a building block that ultimately does fit a real world example, but details to connect it to reality are far beyond the scope of the course, which is why the author will not provide that information. An application for an equation like Mr H's example, comes from a 2nd order differential equation, whose solution is a sum of two exponential functions. A real world example, is a mass/spring/damper, like a car's suspension. There are 3 kinds of solutions to this equation, called underdamping, overdamping, and critical damping. You desire critical damping, as it produces the fastest response time without vibrations. At full capacity, you design for critical damping, which means you get overdamping when its load is less. Here's an example with data: A 2000 kg car (m) has a suspension with a spring stiffness (k) of 81 kN/m, and a damping constant (d) of 27 kN/(m/s). The damper is tuned for its full capacity of 2250 kg, but we're interested in what happens for 2000 kg. Suppose a bump launches it up from its neutral position at 4.5 m/s. How much time will it take to settle within 1 cm of its neutral position? Note: I picked these numbers, so it would simplify as nicely as possible, and be somewhat realistic. Using first principles, you set up this diffEQ: m*y"(t) + d*y'(t) + k*y(t) = 0 2000*y"(t) + 27000*y'(t) + 81000*y(t) = 0 Divide thru by 1000, and it simplifies to: 2*y"(t) + 27*y'(t) + 81*y(t) = 0 The specific solution, for this data is: y(t) = e^(-9/2*t) - e^(-9*t) We want time t when y(t) = 0.01 meters. There are 2 solutions, and the larger solution is of interest, since that's when it settles. 1/100 = e^(-9/2*t) - e^(-9*t) Let E = e^(-9/2*t). Thus: 1/100 = E - E^2 E = 1/2 +/- sqrt(6)/5 Solve for the corresponding t, and pick the larger of the two options: t = - 2/9*ln(1/2 +/- sqrt(6)/5) t = 1.02 seconds
Solving the quadratic by completing the square just creates more work than necessary, and also doesn't make the numbers as easy to work with. This is a nice one to factor (OR you can use the quadratic formula, if you don't feel comfortable with solving it by factoring).
@@herbcruz4697 Disagree, having solved quadratics by completing the square quite a few times, while not finding it more work. The quadratic formula is itself derived by completing the square
@@donsena2013Yes, I'm aware that the Quadratic Formula comes from Completing the Square. Nonetheless, solving this one by completing the square gives you more fractions than necessary.
@@herbcruz4697 Does it ? I never noticed that it did. Using the quadratic formula, you start off with a slightly large fraction. O fact, you don't actually use or create fractions in completing the square. What you do is first to move the constant identifier to the right of the ‘=’ You then add a certain constant to each side of the equation, such that you have a complete square left of the ‘=’ You then take the square root of each side, and the rest is simple algebra.
By multiplying the equation by e^x for both sides e^2x-e^x-12=0 Let e^x=y y^2-y-12=0 (y-4)(y+3)=0 y=4 y=-3 e^x=y=4 Take ln for both sides xln(e)=ln(4) *x~1.4* y=-3=e^x *Rejected*
Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula. A=P*((1+r/n)^(n*t))+x Solving this for (t) (A-x)/P =(1+r/n)^(n*t) At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function. Please help!
There is a tool in Excel called What If analysis, or Goal Seek. In my version, it's under the Data tab on the ribbon, about 80% of the way across the screen. You select a calculated cell, and specify a desired value. Then you select a cell that contains a number, rather than a formula. It will then iterate based on the original value of the input cell, to calculate an input that produces as close as practical to your desired output. It will then use an approximation of Newton's method on your spreadsheet, to iterate to find the desired value. This has no trouble if you are solving for an intersection point between curves. Where it can have trouble, is when the two curves meet at a point of tangency. Where it may never calculate an exact hit, unless it is lucky enough to guess right. For obvious reasons, it will also have trouble if the curves never intersect. It also can have trouble, if there are rounding operations involved.
Your example is possible to solve analytically, and you are correct that you need to use logs to solve it. Given: (A - x)/P = (1 + r/n)^(n * t) Let B = (1 + r/n), and let C = (A - x)/P. The equation becomes: C = B^(n * t) Take the log base B of both sides: log_B (C) = log_B (B^(n*t)) The RHS log cancels out the exponential operation, so this gives us: n*t = log_B (C) Since it's common to not have a built-in general base logarithm function, you usually need to use a change-of-base formula, to put it in a form that you can enter in a calculator or computer. This becomes ln(C)/ln(B): n*t = ln(C)/ln(B) Note: Excel has a general base logarithm, where you enter log(C, B) in this case. If you don't specify the base, it defaults to log base ten with this syntax. The ln(x) function calculates natural log. Divide by n, and we now have an isolated t: t = ln(C)/[n*ln(B)] Recall definitions for B & C, and we have our solution: t = ln((A - x)/P)/[n*ln(1 + r/n)]
e^x-1-12e^-x=0 (√e^x-4√e^-x)(√e^x+3√e^-x) √e^x=4√e^-x dan √e^x=-3√e^-x e^x=16e^-x dan e^x=9e^-x e^x=16/e^x dan e^x=9/e^x e^2x=16 dan e^2x=9 e^x=4 dan e^x=3
![How to Get a Developer Job - Even in This Economy [Full Course]](http://i.ytimg.com/vi/6nz8GXjxiHg/mqdefault.jpg)
You explain complicated problems so simply. Easy to understand!
Thank you for the nice comment 👍
If you wanted to include imaginary solutions, you could write -3 in its polar form (3e^ipi) and then take the natural log of that to get another solution, ln3+ipi
indeed
That’s only the principal branch. You have infinitely many solutions for x by adding multiples of 2pi.
Thank you professor for you sharing your knowledge 👍🏻
Excellent! Congratulations on the clarity of your explanations!
i wish you the best wnbk w5yrt kbire 💘💘
Man I really kind of miss studying algebra and calculus after watching your videos for a few days. Your skills are top notch. 👌 👍
I wish I had YOU as a professor!
You made it look so simple. Thank you.
You’re welcome 😊
Brilliant !!
Great vid, many thanks, Sir!😊
Wow .. thank you sir
Most welcome
Thank You, Professor 🙌🙌🙏🙏
Welcome!
Outstanding. Succinct
Excellent lesson. 👍
Glad you liked it!
Nice one, can you make a video about the gauss-jordan elimination
Will do
Nice vídeo.
What an excellent instructor, clear and showing all the steps and excellent shortcuts. Where were you when I was in college (40 years ago)?
Awesome, you make it look so simple. Thank you!
You are so welcome!
Thank you so much❤
I watched this 2 weeks ago. Still dont understand it fulky, but its like this cloud of uknowing slowly is become more and more clear to see through.
Essentially, it's just a hidden quadratic equation. Let capital E = e^x. This means e^(-x) is the same thing as 1/E.
The given equation, e^x - 12*e^(-x) - 1 = 0, can therefore be rewritten as:
E - 12/E - 1 = 0
Multiply thru by E, to clear the fraction:
E^2 - E - 12 = 0
And solve the quadratic for E.
E = 1/2 +/- sqrt(1/4 - (-12))
E = 4, and E = -3 are the two solutions
We only want the positive solution, since e^x is never negative for real values of x. This means, e^x = 4, so we use natural log to invert the function and find x = ln(4).
Nice problem.
On the last step, you can reexpress ln(4) as 2ln(2) and since ln(2) ~= 0.693 is more familiar to a student and more likely memorized than ln(4), we would not need a calculator.
It's fine to write ln(4) as 2*ln(2).
e^x -1/12e^x -1=0
e^0 -12e^-0 -1=-12
e^1 -12e^1 -1=-2,69
e^1,385 - 12e^-1385 -1=0,009
x=1,385
nice explanation ❤
Thank you~
Thank you very much professor 👍
You are very welcome
I am watching many math solving clips on TH-cam, but as a student my concern was more with; how to I derive such an equation from observing my surrounding or something. That skill is more rewarding than solving it, in my opinion, because there are so many other tools to solve math problems right now. Take this equation for instance. Assuming that an entity has this behavior, how do I observe that and translate this behavior into this/a mathematical format. That's what I have always been looking for in math class but was never taught that in school.
That is a good question. The short answer is to become familiar with typical applications of each of the parent function types, such as linear equations, quadratics, exponentials, and trig. A common application of exponential functions is growth or decay. It's common for situations to either grow or decay, in proportion to the amount you have at any given instant.
Population grows exponentially (at least initially). We could give the number of kids each couple has (suppose 3), and an average age of parents at birth (suppose 30). Since 2 kids replace the parents, this means only one child per woman, contributes to population growth. For this data, every 30 years, the population increases by a factor of 3/2. We can write the equation P(t) = P0 *(3/2)^(t/30), where P0 is the initial population at t=0. You could also write this as P(t) = P0 * e^(t*ln(3/2)/30).
Temperature decays exponentially. Consider cup of tea, initially at 90 Celsius, and you take it outside for your morning walk to work on a cold 0 C morning. After 1 minute, suppose it cools to 80 C. You are interested in its temperature (big T) as a function of time in minutes (little t). This allows us to construct the equation, T(t) = 90*e^(-k*t), where k is an unknown constant. Newton's law of cooling is the physical basis for this equation, which tells us that cooling rates are proportional to temperature difference. Exponential functions are the solution to this constraint. Construct another version, at t=1 min, and we can solve for k. Our function becomes T(t) = 90*e^(-ln(9/8)*t). Maybe you'd like to know when it will be 50C, so it's cold enough to drink; answer: t=5 min.
Not every example in a math textbook, necessarily comes from a real world example. Some come from just curiosity to explore how a function works. It could be a building block that ultimately does fit a real world example, but details to connect it to reality are far beyond the scope of the course, which is why the author will not provide that information.
An application for an equation like Mr H's example, comes from a 2nd order differential equation, whose solution is a sum of two exponential functions. A real world example, is a mass/spring/damper, like a car's suspension. There are 3 kinds of solutions to this equation, called underdamping, overdamping, and critical damping. You desire critical damping, as it produces the fastest response time without vibrations. At full capacity, you design for critical damping, which means you get overdamping when its load is less.
Here's an example with data:
A 2000 kg car (m) has a suspension with a spring stiffness (k) of 81 kN/m, and a damping constant (d) of 27 kN/(m/s). The damper is tuned for its full capacity of 2250 kg, but we're interested in what happens for 2000 kg. Suppose a bump launches it up from its neutral position at 4.5 m/s. How much time will it take to settle within 1 cm of its neutral position? Note: I picked these numbers, so it would simplify as nicely as possible, and be somewhat realistic.
Using first principles, you set up this diffEQ:
m*y"(t) + d*y'(t) + k*y(t) = 0
2000*y"(t) + 27000*y'(t) + 81000*y(t) = 0
Divide thru by 1000, and it simplifies to:
2*y"(t) + 27*y'(t) + 81*y(t) = 0
The specific solution, for this data is:
y(t) = e^(-9/2*t) - e^(-9*t)
We want time t when y(t) = 0.01 meters. There are 2 solutions, and the larger solution is of interest, since that's when it settles.
1/100 = e^(-9/2*t) - e^(-9*t)
Let E = e^(-9/2*t). Thus:
1/100 = E - E^2
E = 1/2 +/- sqrt(6)/5
Solve for the corresponding t, and pick the larger of the two options:
t = - 2/9*ln(1/2 +/- sqrt(6)/5)
t = 1.02 seconds
@@carultch Thank you for the reply and clarity
And e to the x will never be 0. If it were 0=0 , and any x is a solution... an infinite 1. But then the equation does not need be exponential.
Solving the quadratic independently, I completed the square and obtained the same result for X, discarding the negative alternative for X
Solving the quadratic by completing the square just creates more work than necessary, and also doesn't make the numbers as easy to work with. This is a nice one to factor (OR you can use the quadratic formula, if you don't feel comfortable with solving it by factoring).
@@herbcruz4697 Disagree, having solved quadratics by completing the square quite a few times, while not finding it more work.
The quadratic formula is itself derived by completing the square
@@donsena2013Yes, I'm aware that the Quadratic Formula comes from Completing the Square. Nonetheless, solving this one by completing the square gives you more fractions than necessary.
@@herbcruz4697 Does it ? I never noticed that it did. Using the quadratic formula, you start off with a slightly large fraction. O fact, you don't actually use or create fractions in completing the square.
What you do is first to move the constant identifier to the right of the ‘=’
You then add a certain constant to each side of the equation, such that you have a complete square left of the ‘=’
You then take the square root of each side, and the rest is simple algebra.
Exelent
By multiplying the equation by e^x for both sides
e^2x-e^x-12=0
Let e^x=y
y^2-y-12=0
(y-4)(y+3)=0
y=4
y=-3
e^x=y=4
Take ln for both sides
xln(e)=ln(4)
*x~1.4*
y=-3=e^x
*Rejected*
👍
Please make a video to find angle of inverse ratios(sine, cosine....)
.e^x-12e^-x-1=0
Dikalikan dg e^x menjadi
.e^2x-e^x-12=0 yg berarti
.e^x=4 & e^x=-3
.x=log4/loge=ln4
Put directly epoer x euals a .. now its very quick solving
i wish i had youtube back in the day
ex-12e-x-1 = 14 1 12x3=36+3=39 1.39 0 -1 ex 13x3=39 1.39-1.39=0
=(e^x-4)(e^x+3)=0=>e^x={4,}=>x=log4
Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula.
A=P*((1+r/n)^(n*t))+x
Solving this for (t)
(A-x)/P =(1+r/n)^(n*t)
At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function.
Please help!
There is a tool in Excel called What If analysis, or Goal Seek. In my version, it's under the Data tab on the ribbon, about 80% of the way across the screen.
You select a calculated cell, and specify a desired value. Then you select a cell that contains a number, rather than a formula. It will then iterate based on the original value of the input cell, to calculate an input that produces as close as practical to your desired output. It will then use an approximation of Newton's method on your spreadsheet, to iterate to find the desired value.
This has no trouble if you are solving for an intersection point between curves. Where it can have trouble, is when the two curves meet at a point of tangency. Where it may never calculate an exact hit, unless it is lucky enough to guess right. For obvious reasons, it will also have trouble if the curves never intersect. It also can have trouble, if there are rounding operations involved.
Your example is possible to solve analytically, and you are correct that you need to use logs to solve it.
Given:
(A - x)/P = (1 + r/n)^(n * t)
Let B = (1 + r/n), and let C = (A - x)/P. The equation becomes:
C = B^(n * t)
Take the log base B of both sides:
log_B (C) = log_B (B^(n*t))
The RHS log cancels out the exponential operation, so this gives us:
n*t = log_B (C)
Since it's common to not have a built-in general base logarithm function, you usually need to use a change-of-base formula, to put it in a form that you can enter in a calculator or computer. This becomes ln(C)/ln(B):
n*t = ln(C)/ln(B)
Note: Excel has a general base logarithm, where you enter log(C, B) in this case. If you don't specify the base, it defaults to log base ten with this syntax. The ln(x) function calculates natural log.
Divide by n, and we now have an isolated t:
t = ln(C)/[n*ln(B)]
Recall definitions for B & C, and we have our solution:
t = ln((A - x)/P)/[n*ln(1 + r/n)]
@@carultch awesome. Thank you for this. Is the final formula suppose to have a solo "n" denominator? (ln(c)/ln(b))/n?
@@96indashade Yes, I've corrected it.
e^x-1-12e^-x=0
(√e^x-4√e^-x)(√e^x+3√e^-x)
√e^x=4√e^-x dan √e^x=-3√e^-x
e^x=16e^-x dan e^x=9e^-x
e^x=16/e^x dan e^x=9/e^x
e^2x=16 dan e^2x=9
e^x=4 dan e^x=3
e square power 2=e power 9
ex-12e-x-1=0
0=0
math
Thank you so much❤