An Exponential Equation From X
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@yogesh193001 6 күн бұрын
At 6:18, There's a mistake, or am I missing something? 2 * e^2pini =2^1+2pini How can you add the powers when the bases are not the same? Edit: keep watching the video😂😂😂😂😂😂😂😂
@chris999999999999 4 күн бұрын
Boo to the people that demand you edit out all errors. Math is often an iterative process mentally where you back up and try again and not showing that really isn't teaching math.
@SidneiMV 6 күн бұрын
2^x = a (a³ + a²)/(a² - a) = 6 a²(a + 1)/[a(a - 1)] = 6 a(a + 1)/(a - 1) = 6 a² - 5a + 6 = 0 (a - 2)(a - 3) = 0 a = 2^x = 2 => *x = 1* a = 2^x = 3 => *x = (ln3)/ln2*
@mcwulf25 6 күн бұрын
The logic I used was as follows: Cancel 2^x and factorise to get 2^x * / = 6 = 2 * 3 So x=1. But that's just integer solutions!!!
@MaxCubing11 6 күн бұрын
Easy, but when I multiplied the denominator by 6 I did in my head 6×1=1 and wrote one, and for that mistake I got a different solution 😢
@iquisickis2103 3 күн бұрын
At 2:12 I feel like you should make sure that y^2-y is not 0 before cross multiply. It can happen only if y = 0 or y =1. y cannot be 0 and y = 1 can happen if x = 0. Luckily you did not find x =0 as a solution but we must always be careful!
@SyberMath 3 күн бұрын
agreed!
@WahranRai 5 күн бұрын
2:02 WHY DONT YOU SIMPLIFY BY Y the expression before expanding ?!
@SyberMath 3 күн бұрын
Why would you? 🤪
@scottleung9587 6 күн бұрын
Got them easily!
@manwork6545 6 күн бұрын
easy, peasy! have a good we...
@aurochrok634 6 күн бұрын
t = 2^x and we get a simple quadratic. easy.
@SeekingTheLoveThatGodMeans7648 6 күн бұрын
Did I hear that correctly? "Argonne plane"? I would think Cartesian plane.
@phill3986 6 күн бұрын
😊😊😊👍👍👍
@user-yd4ky5vb3w 7 күн бұрын
Thanks for an other video master
@DonEnsley-yi2ql 6 күн бұрын
x ∈ { 1, ln 3/(ln 2) }
@rakenzarnsworld2 6 күн бұрын
x = 1
@lapointe64 5 күн бұрын
why leave that bit in?
@SyberMath 3 күн бұрын
Why not?
@broytingaravsol 7 күн бұрын
so easy
@thexavier666 5 күн бұрын
There is not much difference between the 2 methods
@prollysine 3 күн бұрын
let u=2^x , -> u^3-5u^2+6u=0 , u(u^2-5u+6)=0 , /// u=0 , 2^x=0 , not a solu. /// , u^2-5u+6=0 , u= 3 , 2 , u=3 -> 2^x=3 , x=ln3/ln2 , u=2 -> 2^x=2 , x=ln2/ln2 , x=1 , test , x=ln3/ln2 -> (8^(ln3/ln2)+4^(ln3/ln2))/(4^(ln3/ln2)-2^(ln3/ln2))=36/6 , 36/6=6 , x=1 -> (8+4)/(4-2)=12/2 , 12/2=6 , OK ,
bprp math basics
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