Italy - Math Olympiad Problem | A great approach
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- Опубліковано 15 жов 2024
- You should know this approach. Many goes WRONG!
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More efficient method is to cube both sides and swap exponents on the left to get
27^x=x^27, x=27
Very nice and powerful solution
very nice
@@ricordiaerei7776 3^27=7,625,597,484,987
27^9=13,808,742,332,673
Clearly, the solution is wrong. Use numerical methods to get the answer.
@@bhattajayendra how can that be? 27⁹ = (3³)⁹ = 3²⁷ , chech your sources 27 is a solution
Oh, I missed the calculation. Thanks for the correction.
Lowest approx value for x is 1.1508 by using graphical method
3^x = x^9
On taking log on both sides
x log 3 = 9 log x
log 3/logx = 9/x
We know Log m/ Log n = Log m with base n
Log 3 with base x = 9/x
So, x^(9/x) = 3
Put x=27, it will satisfy.
Umm that's the question?
3^x=x^9 or 3=x^[9/x]
Bro, you can’t just put it inside, you have to show how to have to answer
@@暫時沒有名字-h4z Sorry. If you have a better approach then ki dly share
Ok k here is the question
Put X = 27
It satisfies...soo your answer is 27...
How did you know to rewrite 3^(1/3) as 3^[3/3 * 1/3)? It looks like a stop you only get by playing around, not a step you can get sytematically.
Just math being math. You never know what you have to use and when you have to use it. Thats the beauty of it; it tests your skill and practice
Just use logarithms dude 🤦♂️
True
I think it's olympiad problem for 6th grade
@@COL-eq3rl totally
Show it then, because I don’t think you can
3^x=x^9
Taking log base 3 both sides
X =9.logx
What next?
This equation has two solutions x=1.15082 and x=27. You should specify whether the solution is real number or integer or even complex.
Mentions it at 5.20
Once you have 3^(x / 9) = x, considering other solutions besides natural numbers are not covered in this approach, why not the following:
3^(x / 9) = 9 * (x / 9)
u = x / 9
3^u = 9u = 3^2 * u
3^(u - 2) = u
Trying u = 1, 2, and then 3, the first solution as a natural number gives us u = 3.
x = 9 * u = 9 * 3 = 27.
How? 3^(x/9)=9*(x/9)
@@krishnakalidindi6671 Did you read my entire comment?
@@Limited_Light yes.I could not understand your first step.
@@krishnakalidindi6671 At time 1:44 , Higher Mathematics has the first step, that I began from, 3^(x / 9) = x.
My change in direction uses x = 1 * x = (9 / 9) * x = 9 * (x / 9). (It's multiplication on the right side, not exponentiation.)
Now that the exponent on the left has a similar form to something on the right, I substituted u for x / 9.
It's ok.very good.But it's a trial and error method.Good.
q = 100
for i in range(-q,q+2):
if 3**i == i**9:
print(i)
elif 3**(i+0.5) == (i+0.5)**9:
print(i+0.5)
Cocite dayni
Clever
Outline a plan BEFORE you show a solution. The plan and your thinking are much more important than the grind. In addition, when you use the number one, a vertical line is preferred. In fact, I cant remember any teacher ever using your "1"for 1.
Nothing wrong with his 1. It disambiguates against l or I, as the typeface you're reading now demonstrates
By using numerical method (Newton Raphson) I find the answer is x= 2.485
By calculating This 3^27 is much larger than 27^9
3^x = x^9
Cubing both sides,we get
27^x = x^27
x^1/x = 27^1/27
x = 27
Comparing both sides,we get AA
How cubing both sides will get 27^x ?
(3^x)^3=(3^3)^x=27^x
Also (x^9)^3=x^27
This difficult problem can be solved solely by the application of several laws of exponents. My AI software, MathTeacherXL, could not solve the problem even after I provided the answer! Instead of applying the laws of exponents, it tried using logarithms and approximation techniques, all to no avail.
I used the Lambert W function to solve this question, but I am getting only 1.151 and -0.8963(which is not possible) as solution and not 27, can anybody tell, why?
Thank you.
Can you show your process?
You're wrong LHS is not equal to RHS when you go back to the equation. Your first mistake was saying x/x without putting parameter of that x is not equal to zero
Thanks, very good
If you convert it to A^3 - B^3 and expand, you will have the 2nd equation for the 2nd solution.
Try newton raphson method
Very nice
Using Lambert W method produces 1.1508. Do you know how to use W method to get 27? Is it another branch?
Yes. 27 comes from the -1 branch. There are infinitely many branches but only the -1 and 0 branches produce real results.
3^1/3= 3^3/27 = 27^1/27
When you raised one side to the power 1/3 why didnt you raise the other side?
He also raised to power of three, which cancels out to give one
This is a nice trick for this case - where 3, 9, and 27 are obviously related. Perhaps a more useful approach for two rapidly increasing functions is to take logarithms and compare f₁(x) = (ln3)x with f₂(x) = 9lnx. For x > 0 f₁ is linear and positive, while f₂ is logarithmic, negative with a highly positive slope for small positive x, but positive for x > 1, with a decreasing slope as x grows. That immediately suggests the possibility of two solutions and graphing f₁ and f₂ reveals that easily.
All time ans will be 3*9=27 this is formula
I am not sure about the fact that you reduced by (1/x), before doing that, you must define x≠0
ok plug in x=0 into the given equation. it wont be an equality. this implies that x≠0
Thanks 🙏 sir
Lambert's W function is far too advanced a tool for tackling the non-trivial solution. Here´s my two cents: plot the linear function f(x) = ln3·x and the logarithmic function g(x) = 9·lnx. Obviously, f(1) > g(1), but f(2)
Couldn´t refrain from doing it the N-R way, and the iterating function is not as cumbersome as I had anticipated, since it is t_(n+1) = (t_n - ln t_n) / [1- (ln3 / 9)]. Putting t_0 = 1, it takes exactly as many iterations (13) as before, to reach the same result x = 1,1508248213. Again, a little bit surprised at the rather slow rate of convergence!
your solving method is not complete, if you use Lambert W function there is another solution for x
Go to 5 minutes and 30 seconds he says just that…
3^27=7,625,597,484,987
27^9=13,808,742,332,673
Clearly, the solution is wrong. Use numerical methods to get the answer.
They are both equal to 7,625,597,484,987
You did something wrong
Just take LCM of numbers,
Simple and easy way
When did mathematics reduce to only one, mathematic? I'm old so I may have missed the memo.
3^27=7625597484987
And
27^1÷9=1.44224957031
Both RHS and LHS are not equal
Yes, I firmly agree with that
By using newton Raphson method I find the answer is approximately 2.458
Is there any number else that would have a rule like this
The fact that you're writing "X" like 1 "C" reversed and another normal "C" triggered me. I like it.
everyone does that
X=9/ln3
x=1.15082 and x=27
X=27
log?
😂😂 power of reasoning..
3^(1/3) just entered the chart
Why did the add exponent 3 and exponent ⅓
(x ➖ 3x+3)
Wow, writing 1 like that must get confusing
Simply log would do it quickly
So useful
Meanwhile me and my trial and error solving
😙just use log on both sides
stupid idea , you wil get x=9log3(x) you need to use
derivative to solve it
@@АндрейКузнецов-к3ч1ч If u can use log then u can also use derivative and then integrate it ezz
@user-cg8qe5xx1s please learn to be polite.
Use log and antilog
@@АндрейКузнецов-к3ч1ч It is indeed a great idea. See this approach
3^x = x^9
On taking log on both sides
x log 3 = 9 log x
log 3/logx = 9/x
We know Log m/ Log n = Log m with base n
Log 3 with base x = 9/x
So, x^(9/x) = 3
Put x=27, it will satisfy.
Bro whats the use of this?
Надо ещё показать, что других решений нет.
27
This is the same as your 6 days ago video
x=27
Me who solved it using logarithm
X2= 1.150824
This is the second time you share the same video
What's up with that?
@Math zone jr
Using Lambert W
3^x=x^9
x*ln(3)=9*ln(x)
x^(-1)*ln(x)=(ln(3))/9
-ln(x)*e^(-ln(x))=(-ln(3))/9
case 1:
-ln(x)=W[(-ln(3))/9]
-ln(x)=W[-0,122068]
-ln(x)=-0,140479
x=e^0,140479
=>x=1,1508249
case 2:
-ln(x)=W[(-3*3*ln(3))/(3*9)]
-ln(x)=W[27^(-1)*ln27^(-1)]
-ln(x)=W[ln27^(-1)*e^ln27^(-1)]
-ln(x)=ln27^(-1)
-ln(x)=-ln(27)
=> x=27
Using The Product Log (W) We Get
3^x=x^9
3^(x/9) = x^(9/9)
3^(x/9) = x
( 3^(x/9) ) / ( 3^(x/9) ) = x / ( 3^(x/9) )
1 = x / ( 3^(x/9) )
1 = x * 3^(-x/9)
1 = x * e^((-ln(3)x) / 9)
-ln(3)/9 = ((-ln(3)x) / 9) * e^((-ln(3)x) / 9)
( y = ((-ln(3)x) / 9) )
-ln(3)/9 = ye^y
W(ye^y) = W(-ln(3)/9)
y = W_-1(-ln(3)/9)
y = -3ln(3)
(-ln(3)x) / 9 = -3ln(3)
-ln(3)x = -27ln(3)
x = 27
(There are actually 2 solutions but this is the answer shown in the video)