Solving Exponential Equation With Radicals | Japanese Olympiad Math Question | Mathematics.
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- čas přidán 1. 06. 2023
- In solving this Olympiad Math Exponential Equation with radicals, I will guide you on how to solve for the roots using a different approach different from the normal one you know.
In this Japanese Olympiad question, you will also learn how to handle the radicals with easy, formulate quadratic equation from the radical equation, substitute and factorize quadratic equation with easy.
I will lead you on how to use the natural logarithm to solve exponential equations.
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• Solving Exponential Eq...
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Can u give more questions like this?As the question was amazing and also the way to explain .i enjoyed ur explanation
We are glad you enjoyed and gained something valuable from this video.
Sure, we can make more of this sir.
Thanks for the request.
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I firmly believe that either we should have restrictions for the solutions of the equation or, at least, if we ommit the restrictions we should check the solutions before we accept them.
Thanks for this wonderful comment sir. I have checked the two solutions and they are correct.
I'm proud that I could do this in my mind
Good problem. Good explanation. Thanks for the video lesson!
Thanks for the encouragement sir.
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WOW - THANK YOU.
You're very welcome sir.
We are glad you gained some values from this tutorial sir.
Nice 👍
Thanks sir.
hope see you again
My teacher thanks for everthing
Thanks for making this video sir.
You are welcome always ma.
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Great explanation. This is lovely sir.
Thanks @Daniel Franca
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Excellent videos. Greatly appreciated!
Glad it was helpful and thank you for appreciating our little effort sir.
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@@onlineMathsTV you're very welcome 👍
I think it's very important to check first the existence conditions of some functions, as this square root, to detect non valid solutions or include in your solving
Anyhow you are a beast 😎
Congrats,bro !
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Thank you Sir
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Eu não fazia ideia de pra onde ir 😅😅 vídeo muito bom
😍😍😍 funny you sir.
Coming here means you will definitely know where to go next time you come across any mathematical challenge/problem as u consistently visit OnlinemathsTV channel.
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if you put 3^x =a, 4^x=b you can rewrite the question as this : a-b= root( ab-b^2)
and square each side that can be (a-b)^2 = ab-b^2 ; a^2 - 2ab + b^2 = ab-b^2 ; a^2 - 3ab + b^2 = 0 ; (a-b) (a-2b) =0 ;
a-b=0 or a-2b=0 ; a=b or a=2b ; 3^x= 4^x or 3^x = 2* 4^x ; its more simple what do you think?
Very concise sir. It is a welcome approach too.
Nice work and we have learnt something new from your comment.
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Hmm, I should try a graphical solution method without drawing a proper graph
Sketch a graph of graph of y = 3^x -4^x .
y = (12^x -16^x)^0.5 when x = 0, y=0, certainly. y will always be +ve when x is -ve. When x is positive will generate complex numbers
so, by inspection, both graphs meet at x=0 and y=0
Then try x = -1, -2 and -3 ; gradually the difference in y values get smaller and smaller
In an exam, this sort of question requires an approximate answer for the second value
Выражение под корнем должно быть не меньше нуля. Те, когда возаодили корень в квадрат, должны были получить модуль и нужно рассмотреть 2 случая
I solved the exercise in 2 minutes․ x
Bravo👍👍👍
You the master👌👌👌
@@onlineMathsTV Thank U
thank you very much, i suggest that the value of zero for u should be rejected, for otherwise the division by ( 3 px - 4 px) being equal to zero would not work
But 3 px - 4 px (i.e 1-1) will give us zero, which implies LHS = RHS.
What do you think sir?
In case 1,there is no need to go for' ln' because raised to the power 0 only gives 1
Noted sir.
3^x-4^x=sqrt(12^x-16^x)
y=3^x-4^x
y=sqrty*(sqrt4^x)
=sqrty*2^x
divide both sides by 2^x
1.5^x-2^x=sqrty=y/2^x
sqrty=y/2^x
2^x=y/sqrty=sqrty
square both sides
y=4^x=3^x-4^x
+4^x both sides
2(4^x)=4^(x+0.5)=3^x
z=ln4/ln3 [3^z=4]
(3^z)^(x+.5)=3^x
x=zx+0.5z
*x both sides
x^2=zx^2+.5zx
-x^2 both sides
(z-1)x^2+0.5zx=0
[1 rough quadratic formula later]
x=(-0.5(ln4/ln3))/((ln4/ln3)-1)
=-ln4/((2ln3)((ln4/ln3)-1)
(2ln3)((ln4/ln3)-1)=((2ln3*ln4)/ln3)-2ln3
=2ln4-2ln3
=>x=-ln4/(2(ln4-ln3))
…and also x=0.
edit: oh wow, different fractions yet the same constant
Wonderful and unique move.
You are good at it sir, you the master here.
Respect👌👌👌
You must set 3^x--4^x>=0 ...... x
noted sir.
Thanks master.🙏🙏🙏
Only one to notice. Hats off
how about 0 or 1/{ log2 (3) - 2}
Thanks for the encouragement sir.
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This also has the advantage that many calculators natively perform lg2, so x = 1 / (lg2(3) -2) is trivial to calculate. In RPN: 3 ; lg2 ; 2 ; - ; 1/x. {five keystrokes}
It seems that you like solving children' s math-probs
3^x - 4^x = sqrt(14^x -16^x). Notice that the expression inside the radical can not be negative, but 14^x < 16^x ---> 14^x - 16^x will always be negative. There is only 1 way to neutralise the negative result is to make the expression= O. since any value taken to pwer of 0 will become 1. If x = 0, 24^0 -16^0 = 1 - 1 = 0
Checking whether x = 0 the solution:
3^0 - 4^0 = sqrt(14^0 - 16^ 0)
1 -1 = sqrt(1 - 1)
0 = 0. Hence, x = 0 is the solution
We're done
Ok sir, thanks for dropping this sir.
We learn everyday from the best brains all around the globe and we sincerely believe you are one of these rare genius in mathematics. Nice to have you here sir,
You are most welcome master.
X не может быть равен "0", так как -1 не равно sqrt(-1)
If we use log instead of using ln ,then what will happen
same result sir.
Using logs to the base of 10, we can fill in the numbers (the logarithms of 2, 3 and 4, which are 0.3010, 0.4771 and 0.6020) from memory and then calculate the quotient by hand. I don't have natural logs in memory.
I tried my Genetic Algorithm with this; 'x' must be a negative number, and it needn't be an integer. There seems to be no real solution, the greater the (negative) number, the closer it gets. I scanned from 0-> -500 and the best I got was:
X = -481.58408788993336
Left side evaluates to: 1.6826575283911937e-230
Right side evaluates to: zero (probably not real zero, the computer just doesn't display such a small number)
Difference = 1.6826575283911937e-230 , effectively solving the equation (for engineers, not theoreticians!)
(plus overflow errors and other Python squawking)
Maybe I should watch the video ......
.
Smiles....nice work from you boss.
But most times AI fails to give the actual result to a mathematical challenge.
I believe you have watched the video sir.
Did you agree with the approach I applied here?
I made a video on a similar challenge with a different approach.
Below is the link to that video, you can as well flip through it to check it out if you don't mind sir.
czcams.com/video/0cWjl5616pQ/video.html
And do not hesitate to make corrections where necessary because we are here to also learn from the best brains like you all around the world.
Thanking in advance boss!!!
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Do you mean "generic" algorithm?
I disagree with x2 solition - it is superfluous as we should use straight definition of logarithm: X2=log(3/4)2. log(a)b.
not too clear with your point here sir.
What's is the meaning of Wn( 2) ?????
Wn(z) is generally called the Lambert W function (or omega function or productlog) where the subscript n (an integer) represents the principal, positive or negative branch of the function with n = 0 being the principal branch.
The Lambert W function is the inverse of the function:
f(w) = we^w
where w is any complex number and e^w is the exponential function.
The function can also be written as:
LambertW(n,z)
ProductLog(n,z)
,X= log( 3/4 ) 2
×=-3
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Hello students
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X1=0
X2~ -2.41
Bravo 👍👍👍
You the best sir.
ln 1 is equal 0 so x1=0
Yap
Easier to go m log7=log70
Bravo 👍👍👍
You the best sir.
In this video, the respected teacher is attempting to divide by zero, the value for which remains undefined and unknown. Thereby, his method is wrong.
Sir, are you of the opinion that zero as a root to this math challenge is wrong? Kindly do the simple substitution and check if an undefined expression will emerge in this math challenge sir.
Thanking you for a positive response in advance.
I think there is no division by zero. In [(3^x)-(4^x)]²=(4^x)[(3^x)-(4^x)] the teacher didn't divide the equation by (3^x)-(4^x), rather he let (3^x)-(4^x)=u and then move all terms to one side.
Directly moving all terms to one side follow by factorizing we get
[(3^x)-(4^x)][{(3^x)-(4^x)}-(4^x)]=0
[(3^x)-(4^x)][(3^x)-2(4^x)]=0
Hence: • (3^x)-(4^x)=0 --> x=0
• (3^x)-2(4^x)=0 --> (¾)^x=2
NO DIVISION BY ZERO
빙빙 돌아가며 풀이해 지친다..
무조건 자연로그로만 푸니 듣기가 힘듬
You don't have to copy all such lengthy steps of human -invented motions of algebra to arrive at the correct answer
Again, logical reasoning can help you solve all sorts of problems, but not copying such lengthy formulas and steps, like a photocopying machine
Noted and thanks sir.
X=0
Correct boss, but that is just one of the solutions.
Thanks for stopping by sir.
Much love.
I already see 0 as solution 😅
Yeah but thats only one solution!!
I wonder who is meant as the audience of this piece. A general mediocre pupil will not get even this long and overworded explanation. A bit developed one (1) does not need 2/3 of this wording, (2) guesses x=0 is a solution in the first 30 seconds, by substitution, (3) since a square root from a negative number is not real, gets any other root is negative. Finaly,
3^x = 4^(x+1/2) can be trivially solved with much less wording.