Solving An Interesting Log Equation | Math Olympiads
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- เผยแพร่เมื่อ 30 ก.ย. 2024
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Nice - it's fun watching you do this one, and I'm glad I didn't even think about attempting it!
Good question, good solution. I think x = exp(a^(a/(a-1))), where a = ln(10), is another way to write the same solution, but using only one exp function.
indeed!
10^(a^(1/(a-1)))
Where a is ln10
Let y=log x = ln x / ln 10
Then ky = ln x
where k=ln 10
y^(ky)=(ky)^y
(y^y)^k=k^y * y^y
(y^y)^(k-1)=k^y
ln both sides gives
y * (k-1) * ln y = y * ln k
provided y is not 0
we have ln y / ln k = 1/(k-1)
logk(y) = 1/(k-1)
y = k^(1/(k-1))
x = 10^y
x=10^(k^(1/(k-1)))
I thought that the plot was y=ln(x)^log(x) and y=log(x)^ln(x) that should be interesting
Yes, I was hoping to see that. I'd expect to see two intersection points, where the one at x = 1 would be a "hole" in the curves.
謝謝！
Thank you!!! 🤩🥰🤩
Its look like this equation Xʸ=Yˣ
Compairing you got y=lnx=4
x=54
But why its is not like your solution!?
This equation has 2 more answers, X = e , X = 10 are also the answers.
No, the equation is (log x)^(ln x) = (ln x)^(log x)
(log 10)^(ln 10) = 1^(ln 10) = 1
(ln 10)^(log 10) = (ln 10)^1 = ln 10
1 =/= ln 10 so x = 10 is not a solution
(log e)^(ln e) = (log e)^1 = log e
(ln e)^(log e) = 1^(log e) = 1
log e =/= 1 so x = e is not a solution
Me immediately: 1
Sorry, the justification for why 0^0=0^0 invalidates the root x=1, does not convince me.
0^0=1
I just graphed the function, and exactly, x=1 is a root.
0^0 is not 1, it's undefined, it's the same as 0^(1-1)=0/0
@@unonovezero indeterminate =/= undefined
0^0 is not 1, it is indeterminate
1. Your graph only suggests that lim(x->0) x^x = 1 and even though this limit is true, it doesn't say that 0^0 = 1
2. f(x) = 1/x and g(x) = e^x / x^2, you wouldn't say that f(0) = g(0) since they both are 1/0, right? Cuz that would be wrong too.
@@Packerfan130 Dude, in algebra and calculus, 0^0=1, yes, why? Because it has coherence with all theorems, properties, and results. In algebra and calculus. That is explained in all the universities, 0^0, it is indeterminate when it comes from a limit as such. But if you are given the expression 0^0, without coming from any function, it is 1
At time 10:58 you said the graph was a horizontal line. Didn’t you mean it’s a vertical line?
I used the same approach! I am glad I solved it❤❤❤
Nice :)
Thanks!
OMG its Albert Einstein....Its Albert Einstein OMG
solution starts at 4:58
Background info starts at 0:00 for those who are not that advanced 😁😲😜
If you ln both sides the domain needs to be x>=1. Was the domain given in the original problem?
No. You need to find it 😁😜
x=e^(1/(1-1/loge))
Crazy😃💯💯💥
"Even better than the real thing" ;)
😍
Unfortunelly, I couldn't
No worries. Did you see my solution?
@@SyberMath Yep
Respectfully, this presentation was all over the place.
Easy!!!
To solve this equation, we can start by taking the natural logarithm of both sides:
ln((log x)^ln x) = ln((ln x)^log x)
Using the property of logarithms that ln(a^b) = b ln(a), we can simplify the left-hand side:
ln(log x) * ln x = ln(ln x) * log x
Next, we can use the change-of-base formula for logarithms to convert the logarithms to a common base, say base 2:
(ln(log x) / ln(2)) * ln x = (ln(ln x) / ln(2)) * log x
Simplifying this expression by multiplying both sides by ln(2) and dividing by log x and ln x, we obtain:
(ln(log x) / log x) = (ln(ln x) / ln x)
Now, we can use the fact that f(x) = ln x / x is a decreasing function for x > e (where e is the Euler's number) to conclude that both sides of the equation are equal only if log x = ln x. However, this equation does not have a closed-form solution, so we must resort to numerical methods to approximate its solution. One such method is to use iterative techniques, such as Newton's method or the bisection method, to find a root of the function f(x) = ln x - log x.
log x = ln x does have a closed form solution, though; it's just x = 1. However, in your last step it seems like you've made an assumption that eliminates the other solution :)
(logx)^lnx=(lnx)^logx
Let logx=a
lnx=b
a^b=b^a
a=b
logx=lnx
x=1
the other solutions are a=t^(1/t-1)
and b=at
logx=t^(1/t-1)
lnx=logxt
lnx=lnxt/ln10
ln10=t
lnx=logxt
=t^(t/t-1)
lnx=ln10^(ln10/ln10-1)
x=e^(ln10^(ln10/ln10-1))
x=e^(ln10^(1+1/ln10-1))
x=e^((ln10)^(1)*ln10^(1/ln10-1))
x=(e^ln10)^(ln10^1/ln10-1))
x=10^(ln10^(1/ln10-1))
You are missing grouping symbols in many places in your alleged correct work:
[log(x)]^ln(x) = [ln(x)]^log(x)
Let log(x) = a
ln(x) = b
a^b = b^a
a = b
log(x) = ln(x)
x=1
the other solutions are a = t^[1/(t - 1)]
and b = at
log(x) = t^[1/(t - 1)]
ln(x) = log(xt)
ln(x) = ln(xt)/ln(10)
ln(10) = t
ln(x) = log(xt)
= t^[t/(t - 1)]
ln(x) = ln(10)^[ln(10)/(ln(10) - 1)]
x = e^{ln(10)^[ln(10)/(ln(10) - 1)]}
and so on with more steps until
x = 10^{ln(10)^[(1/[ln(10) - 1]}
As we can see in the final answer, it would be better if you write the 1 as it is displayed here : with a little end at upper left, not as an uppercase i. In France, we write 1 as 1, not as I (uppercase i) nor as l (lowercase L). So, when we write ln, the lowercase L is not as the 1, and we can read ln 10, not ln I0. In fact, by handwriting, we use ℓ instead of l. So at 10:49, we don't have ln I0-I. But better ℓn 10 -1 which is more readable. Thanks. I hope you will soon have better teachers in elementary school in the USA, which learns your kids to write correctly (to avoid confusion between 1 and 7 in handwriting, in France, we add a little horizontal bar in the middle. As you have a heavy weight issue in the general population in the USA, a little more work when you write 1 and 7 car help you losing fat.
"learns your kids" It is "teaches your kids." The kids *learn* from the teacher and/or the material. Please don't comment about the weight of the general population in the USA, as it is
irrelevant to the topic, and making the claim about writing those numbers that way in regards to weight is absurd as well.
It started off fine but really took a turn towards the end. Try to be more civil and respectful of other nations. Also, it is "who teach your kids to write correctly" and not "learns your kids". Also I am so sorry for your loss. Being french is the second worst thing to happen to a person after being dead.
Once again, a good 4 minute video... unfortunately it lasted 11 minutes...
My bad! 😜😲😁
@@SyberMath Just a style difference! Good content...