A Great Exponential Equation

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Опубликовано:

27 июл 2024

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Комментарии : 39

@cav1928 17 дней назад

3^x /(2x+1)=1 -->x*ln(3) - ln(2x+1) =0-->x*ln(3)=ln(2x+1)--> by check and guess is easy to find out that x=1 and x=0 are the only solutions. Of course your application of Lambert function, mainly in the right side, is super.

@SyberMath 17 дней назад

good thinking!

@pukulu 16 дней назад

By inspection, the 2 solutions among the real numbers are x = 0 and x = 1.

@alphastar5626 18 дней назад

0 and 1

@nasrullahhusnan2289 17 дней назад

Right, by inspection x={0,1}

@lazymello6778 15 дней назад

But why was the soln x=1 missed in this approach?

@DonEnsley-mathdrum 17 дней назад

x = 0 and 1. The line g(x) = 2x + 1 has slope 2 a constant. The exponential f(x)=3^x has slope ln 3 3^x which is always positive, and increasing with increasing x and so the exponential never has zero slope, and so a line can possibly have either 0, 1 (at a tangent point), or 2 points at most intersections with the simple exponential 3^x. In our case we get the two points only. They are (0,1) and (1,3).

@stephensu4371 18 дней назад

7:20 i’m not sure i’m correct or not, did we missing something on the right hand side? edit: 8:46 ok. fixed

@CryToTheAngelz 17 дней назад

Using the same method, how can we get the second result? I am guessing there is some kind of branching involved but how does it work?

@minecraftexplorer56 10 дней назад

It is also possible to find roots with graphical method.

@Penfold42 17 дней назад

When are you launching the te^t channel ?

@SyberMath 17 дней назад

😁

@moeberry8226 17 дней назад

In this specific problem the Lambert W function is not necessary since you still have to check which function lies above the other over different intervals because they are both increasing but obviously 3^x is increasing much faster after x=1 however before that between 0 and 1 the function 2x+1 is lying above 3^x. But also you haven’t shown between-1/2 and 0 where 2x+1 is still positive and it could possibly intersect at a negative value. This requires more of real analysis and using IVT and Newtons method assuming you have a good initial guess close to the real zero then that will do the job.

@nackha1 17 дней назад

Nice explanation 👍 Just wondering, what software do you use to write on your videos?

@SyberMath 17 дней назад

Notability

@ginopaperino2608 17 дней назад

I have a question. What if i derive both sides. The result shouldn’t change but it does. Is there any situation in wich I can do that?

@maxhenderson1890 13 дней назад

Show me what you've done please

@ginopaperino2608 13 дней назад

@@maxhenderson1890 the derivative of 3 to the x power is 3 to the x power multiplied by ln(3) and the derivative of 2x + 1 is equal to 2. So x should be equal to log3[2/ln(3)] right? i think that we can not derivate both sides of an equation like that. Do u know what is the mistake?

@nasrullahhusnan2289 17 дней назад

To get x=0, it is simpler this way: At 3:34 you get (-x-½)[3^(-x)][3^(,-½)=-½[3^(-½)] It can be simplified as (-x-½)[3^(-x-½)]=-½[3^(-½)] which implies that -x-½=-½ --> x=0 If fact, by inspection it can be found that x={0,1}. I consider your solution is an algebrai way to get x=0. Is there algebraic way to get x=1 instead just by inspection?

@SyberMath 17 дней назад

yes but you need to manipulate the expression to fit the te^t pattern

@el-mehdibenchaib9950 17 дней назад

I solve it and find x=1. I enter the exp on both sides Exp(3^x)=exp(2x+1) Then, exp(3x)=exp(2x+1) Therefore, 3x=2x+1 So we have two solutions (0,1).

@appybane8481 17 дней назад

but exp(3^x) isn't exp(3x)

@zegrirsaid2855 16 дней назад

0and1

@rakenzarnsworld2 17 дней назад

x = 0 or 1

@Chrisoikmath_ 18 дней назад

I have two different aproaches: 1. We can easily find that 0 and 1 are two roots of the function f(x) = 3^x - 2x - 1. We hope that these are the only ones. Let's prove it: It's f'(x) = ln3.3^x - 2 Obviously the function f' has exactly one solution. Let us assume that the function f has three solutions: 0, 1 and t. By Rolle's theorem at [0,1] we have that the only root of f' belongs in (0,1). We can check three cases: A) t1 then by Rolle's theorem at [1,t] there exists a number w€(1,t) such that f'(w) =0, contradiction. 2. Since f' has exactly one solution let it be u then f is decreasing at (-infinity, u] and increasing at [u, +infinity) so f has at maximun two solutions one in (-infinity, u] and one at (-infinity, u]. Because 0 and 1 are roots of f these are the only ones. Sorry for my bad English and thanks for reading.

@forcelifeforce 18 дней назад

Do not use a decimal point for multiplication on your fourth line. It reads as "f(x) = the natural logarithm of 3 point 3 times x^2 - 2." Also, you are missing grouping symbols around the argument. Instead, write f'(x) = ln(3)*3^x - 2, ln(3)[3^x] - 2, or the equivalent.

@saishashank 18 дней назад

A fan of you from india

@SyberMath 17 дней назад

😍 🇮🇳

@rorydaulton6858 18 дней назад

I solved it by using guess-and-check to quickly find the solutions x = 0 and x = 1. But guess-and-check alone cannot show that there are no other solutions. I proved the lack of other solutions by considering the function 3^x. The second derivative is (ln 3)^2 * 3^x which is always positive. That means that the graph of 3^x is concave upward everywhere: it is a convex function. Such a function can have at most two intersections with a straight line. But the graph of 2x+1 is a straight line. Therefore there can be at most two points where the graphs of 3^x and 2x+1 intersect, meaning there are at most two solutions to the equation 3^x = 2x+1. And we already found those two solutions. I checked that conclusion by considering the Lambert W function, as you did. It was clear from the start that your equation could be solved by Lambert W, and we know that there are only two real branches of that function. Therefore there can be at most two solutions to such an equation, which we already found. Your method did this train of thought in more detail.

@saishashank 18 дней назад

Nice 👌

@saishashank 18 дней назад

Where are you from?

@rorydaulton6858 18 дней назад

@@saishashank I'm not sure what you mean. I was born and raised in the United States of America and am living there now but served as a missionary in East Africa (Kenya and Tanzania) for over 25 years. I now live in central Pennsylvania, teaching high school science, calculus, and statistics in a private Christian school. My degree is in mathematics: magna cum laude from Cornell University. Did I cover what you meant?

@saishashank 18 дней назад

@@rorydaulton6858 very interesting and excellent bio... Actually I'm not much experienced as you sir... I'm about to join in a degree "Bachelors of Mathematics" in ISI (Indian Statistical Institute) My age is less than your missionary period. I'm 19 I'm very much interested in mathematics sir.. Glad to meet you sir Any suggestions for me from your side sir??

@rorydaulton6858 18 дней назад

@@saishashank I do not know much about Indian schools so I cannot say much. I believe that you will cover more math than I did, since my degree was Bachelor of Arts which requires many non-math classes. I also knew that I might become a missionary so I took two more missionary-related classes and fewer math classes than I could have taken. I advise you to pay attention to your classes, not specialize yet but figure out which kind of mathematics you like so you can specialize later, and push yourself a little harder than you believe you can handle. And keep learning outside school: I covered convex functions and sets in a couple of my classes but learned the Lambert W function on my own later. Finally, pay attention to the non-academic areas of your life, especially the spiritual. What do you think about Jesus of Nazareth? Trusting and following Him is more important than math. My math and other grades went up after I trusted in Jesus and began student ministry. But that is not the reason to become a committed Christian: there are many better ones, including the need to follow the truth.