While it's trivial, there is a bit of a method for a broader case. Right hand side is a power of 2, therefore x can be written as 2^y. Replace in left hand side, simplify, make powers of 2 equal, it's easy.
To solve this question, knowing 2⁹=512 is enough. Then; x^(3^x)=2⁹ If x=2 so 3^x=9 Condition and result both have the same answer. In such problems, since the two sides of the equality are not intertwined, the condition-result theorem can be used. For example; xe^(x)=-1/e If x=-1 so e^x=1/e
I did it a little simpler way than Method One, by taking logarithms to the base 2 instead of natural logs to the base e. Remember that log2 (2) =1, and log2 (512) = 9 (because 512 = 2^9). So the equation reduces to 3^x = 9. Now take logs to the base *3*, keeping in mind that log3(3) = 1 and log3 (9) = 2 (because 3^2 = 9). 3^x = 9 reduces to x = 2.
We know that x is an even number because 512 is an even number. 4^3^4 is too large but 2^3^2 is exactly right. For completeness, we also know that x is a positive number because the result is an integer (-2)^3^(-2) is not an integer (and 0^3^0 = 0^1 = 0)
While this equation happened to be something that I could very quickly visually solve by knowing the exponents of 2, but I enjoyed seeing how I could approach this type of question with numbers larger/more complicated than this (also when approaching this question before watching the video I completely overlooked checking if it has other solutions). Thank you for the great video
Devoted follower here, really enjoy your channel. A question on procedure. The reason you need to check if 2 is the only solution, is because you couldn’t solve explicitly for X, correct? You came across that 2^9 was equal to 512….and were able to manipulate the equation to an equality, which is why you needed to check to make sure the solution was unique. However, if you were able to solve explicitly for X, you would not need to check, correct?
Solve the equation Davide masi find the equation of circle inscribed in a triangle whose sides are lines line one x+y=8 line two 2x+y=22 line three 3x+y=22
To answer this question, we need to find the intersection of the bisectors of the triangle, which is the center of the circle. Well, for this we need to use the property of the bisector, every point on the bisector is an angle from two sides of the same angle at the same distance. we take. Then, using the formula for the distance of the point from the line, we calculate the distance of the center of the circle from each line. L1[x+y=8] L2[2x+y=22] L3[3x+y=22] |3a+b-22|/(sqrt(10)=R |a+b-8|/(sqrt(2)=R |2a+b-22|/(sqrt(5)=R So {Angle bisector L2L3 } |(3a+b-22)/(2a+b-22)|=sqrt(2) |1+a/(2a+b-22)|=sqrt(2) and {angle bisector L1L3} |(3a+b-22)/(a+b-8)|=sqrt(5) |1+(2a-14)/(a+b-8)|=sqrt(5) To get rid of the absolute value, either you have to increase both sides of the equality to the power of 2 and... or identify the correct signs by specifying the vertices and drawing the triangle. So angle bisector L2L3 a/(2a+b-22)=-sqrt(2)-1 -2a/(2a+b-22)=2(sqrt(2)+1) -a/(b-22)=(sqrt(2)+1)/(3+2sqrt(2)) -a/(b-22)=(sqrt(2)+1)/(sqrt(2)+1)² a/(b-22)=-1/(sqrt(2)+1) a=(22-b)/((sqrt(2)+1){Relationship 1} Also angle bisector L1L3 (2a-14)/(a+b-8)=sqrt(5)-1 (a-7)/(a+b-8)=(sqrt(5)-1)/2 (a-7)/(b-1)=(sqrt(5)-1)/(3-sqrt(5) (a-7)/(b-1)=(sqrt(5)-1)(3+sqrt(5)/4 (a-7)/(b-1)=(sqrt(5)+1)/2 According to relation 1 (22-b)/(sqrt(2)+1)-7)/(b-1)=(sqrt(5)+1)/2 (15-7sqrt(2)-b)/(b-1)(sqrt(2)+1)=(sqrt(5)+1)/2 (15-7sqrt(2)-b)/(b-1)=(sqrt(5)+1)(sqrt(2)+1)/2 (2-sqrt(2))/(b-1)=((sqrt(5)+1)(sqrt(2)+1)+2)/14 (b-1)=14(2-sqrt(2))/((sqrt(5)+1)(sqrt(2)+1)+2) b=(14(2-sqrt(2))+(sqrt(5)+1)(sqrt(2)+1)+2)/((sqrt(5)+1)(sqrt(2)+1)+2) b=(31-13sqrt(2)+sqrt(10)+sqrt(5))/(3+sqrt(5)+sqrt(10)+sqrt(2))~1.84 And with placement 《b》in relationship 1 we have ; a=(22-b)/(sqrt(2)+1) a=(22-(1.84))/(sqrt(2)+1)~8.36 R²=(1.84+8.36-8)²/2=2.42 and the equation of the circle (X-8.36)²+(y-1.84)²=2.42
could not use power rules to rewrtie as x to x to the 3rd power, then take the 3rd root both sides and end with x to x is a number which will be easier to solve since the other side will be
does anyone know how to do Sigma, from n=1 to 2k of (-1)^n+1 * (2n-1)^2 ??? Wolfarm alpha can do it and its = 1/2 (-16 (-1)^(2 k) k^2 + (-1)^(2 k) - 1) but I couldn't find the explanation as to why this is the answer I'd really appreciate if someone is educated on the subject and can help me out
You can go by induction or calculate it explicitly. In the second case you should find out that the sum is equal to: Sum form j=1 to k of {(4j-3)^2-(4j-1)^2} Notice that the summatory changed from 2k to k since i compute two terms every time. I write The propg of this statement at the end of the comment. Then you have (after some calculation) Sum from j=1 to k of {-2(8j-4) } that leads to -8* sum from j=1 to k of (2j-1) It is known the the sum of odd sequential numbers is k^2 so you have finally -8k^2 Demonstration of the first fact: Splitting the sum when n is odd and even you have: First case: n=2j-1 Sum from n=1 to 2k-1 of {(-1)^(n+1)(2n-1)^2} that leads to Sum form j=1 to k of {(-1)^(2j) (4j-3)^2 (Notice that -1 is elevated to an even power so is always 1) Second case: n=2j Sum from n=2 to 2k of {(-1)^(n+1)(2n-1)^2} that leads to Sum form j=1 to k of {(-1)^(2j+1) (4j-1)^2 (Notice that -1 is elevated to an odd power so is always -1) Than you can sum the two summatories with index j and have the result i wrote above.
@@il_caos_deterministico Sorry I should have asked for something else, How do you simplify Sigma from n=1 to any number we'll call it k. Of (-1)^n+1*(2n-1)^2 I'm really sorry Ive wasted your time by not making my question clear..
@@mariothethird5624 it’s not clear still.. i mean, the sum from n=1 to 2k is simply -8k^2, and i have given you how to compute it in the first comment.. you can’t really do something different from what i’ve done (separatong the sum in two parts, reindexing and put them together again)
@@il_caos_deterministico Well you see, I'm still in high school and Im not even supposed to learn these stuff yet but I'm really curious and want to learn about it.. what I'm trying to say is that I don't even know how to do a general simplification of a sigma going from n=1 to k of any sort of sigma and what I'm asking is do you know what it means to simplify a sigma going from n=something to k (meaning a general number and not a specific number)?? And I dont want to waste your time but if you could just direct me to the right place/tell me what to search on google/youtube to learn about it I would really appreciate it. Btw thank you for even replying in the first place that's very nice of you so thanks.
Basically you have to find the equations of two bisectors and intersect them to find the center of the circle, then just use the fact I wrote in the other comment to find the radius R. Then remember that R²=(x0)²+(y0)²-c to find c and complete the equation which is: x²+y²-2x0x-2y0y+c=0
Find the 3 intersections A (xA,yA), B(xB,yB) , C (xC,yC) of the 3 lines : that is your triangle Compute the length of sides AB=c, AC=b, BC=a Recall the incenter I (xI,yI) (the center of inscribed circle in the triangle) is the *centroid of A, B, C with weights a,b,c* ---> a*IA + b*IB + c*IC = 0 ---> projection on oX, oY etc... Use of Heron formula to have the Area of triangle ABC: Area = sqrt(p*(p-a)*(p-b)*(p-c)) with p = 0.5*(a+b+c) then the radius r = Area/p -----> equation of inscribed circle : *(x-xI)^2 + (y-yI)^2 = r^2*
x^{3^x}=512; x should be > 0; x^{3^x}=2^9; if x=2 we have 2^{3^2}=2^9, so x=2 is a solution. Is it the only ? Yes because function x^{3^x} is strictely increasing obviously. Not a very interesting equation.
Simply rewrite 512 as 2^9 and 9 as 3^2 to find that x = 2
you still need to prove that x = 2 is an unique solution to the equation by proving f(x)=x^(3^x) is always increasing
That was quite ready to do just by looking at it .
But it's interesting to know how to do it algebraically
Yes I saw that right away - he said some people would in the video.
Can you prove its the only solution?
-I just looked at the graph in desmos to see it was always increasing-
@@mcwulf25 3^x is always increasing and x is always increasing therefore x^(3^x) is always increasing which means it has one solution only.
While it's trivial, there is a bit of a method for a broader case. Right hand side is a power of 2, therefore x can be written as 2^y. Replace in left hand side, simplify, make powers of 2 equal, it's easy.
Maths students should have maybe the first 10 powers of 2 in their heads. Computer science students should have the first 16.
And anyone who's spent a significant amount of time playing 2048 will just know all those powers. 😊
@@CousinoMacul I spent too much time playing 2048...
To solve this question, knowing 2⁹=512 is enough.
Then; x^(3^x)=2⁹
If x=2 so 3^x=9
Condition and result both have the same answer.
In such problems, since the two sides of the equality are not intertwined, the condition-result theorem can be used.
For example; xe^(x)=-1/e If x=-1 so e^x=1/e
I didn't aim to reply to you specifically, it's a bad smartphone manipulation from me. Sorry.
@@UneFenetreSurLeMonde ok, different analysis can be done but in the end you will arrive at the same answer
never been great at math . your videos are helping me out alot
thanks
Np. Glad to heart that
Nobody:
People who played 2048 a lot: That's too easy
3x=9 => x=9/3 = 3 try this
I did it a little simpler way than Method One, by taking logarithms to the base 2 instead of natural logs to the base e. Remember that log2 (2) =1, and log2 (512) = 9 (because 512 = 2^9). So the equation reduces to 3^x = 9. Now take logs to the base *3*, keeping in mind that log3(3) = 1 and log3 (9) = 2 (because 3^2 = 9). 3^x = 9 reduces to x = 2.
We know that x is an even number because 512 is an even number. 4^3^4 is too large but 2^3^2 is exactly right.
For completeness, we also know that x is a positive number because the result is an integer (-2)^3^(-2) is not an integer (and 0^3^0 = 0^1 = 0)
Good problem! Thanks for the explanation 😁
Np. Thanks
While this equation happened to be something that I could very quickly visually solve by knowing the exponents of 2, but I enjoyed seeing how I could approach this type of question with numbers larger/more complicated than this (also when approaching this question before watching the video I completely overlooked checking if it has other solutions). Thank you for the great video
The answer becomes easily apparent if you factor 512. 512 = 2^9 = 2^3^2, that's how I stumbled onto the answer.
Devoted follower here, really enjoy your channel. A question on procedure. The reason you need to check if 2 is the only solution, is because you couldn’t solve explicitly for X, correct? You came across that 2^9 was equal to 512….and were able to manipulate the equation to an equality, which is why you needed to check to make sure the solution was unique. However, if you were able to solve explicitly for X, you would not need to check, correct?
Solve the equation Davide masi find the equation of circle inscribed in a triangle whose sides are lines line one x+y=8 line two 2x+y=22 line three 3x+y=22
Read my answer in your previous comment. Involve yourself and search google you will find what you are looking for
To answer this question, we need to find the intersection of the bisectors of the triangle, which is the center of the circle. Well, for this we need to use the property of the bisector, every point on the bisector is an angle from two sides of the same angle at the same distance. we take. Then, using the formula for the distance of the point from the line, we calculate the distance of the center of the circle from each line.
L1[x+y=8]
L2[2x+y=22]
L3[3x+y=22]
|3a+b-22|/(sqrt(10)=R
|a+b-8|/(sqrt(2)=R
|2a+b-22|/(sqrt(5)=R
So {Angle bisector L2L3 }
|(3a+b-22)/(2a+b-22)|=sqrt(2)
|1+a/(2a+b-22)|=sqrt(2)
and {angle bisector L1L3}
|(3a+b-22)/(a+b-8)|=sqrt(5)
|1+(2a-14)/(a+b-8)|=sqrt(5)
To get rid of the absolute value, either you have to increase both sides of the equality to the power of 2 and... or identify the correct signs by specifying the vertices and drawing the triangle.
So angle bisector L2L3
a/(2a+b-22)=-sqrt(2)-1
-2a/(2a+b-22)=2(sqrt(2)+1)
-a/(b-22)=(sqrt(2)+1)/(3+2sqrt(2))
-a/(b-22)=(sqrt(2)+1)/(sqrt(2)+1)²
a/(b-22)=-1/(sqrt(2)+1)
a=(22-b)/((sqrt(2)+1){Relationship 1}
Also angle bisector L1L3
(2a-14)/(a+b-8)=sqrt(5)-1
(a-7)/(a+b-8)=(sqrt(5)-1)/2
(a-7)/(b-1)=(sqrt(5)-1)/(3-sqrt(5)
(a-7)/(b-1)=(sqrt(5)-1)(3+sqrt(5)/4
(a-7)/(b-1)=(sqrt(5)+1)/2
According to relation 1
(22-b)/(sqrt(2)+1)-7)/(b-1)=(sqrt(5)+1)/2
(15-7sqrt(2)-b)/(b-1)(sqrt(2)+1)=(sqrt(5)+1)/2
(15-7sqrt(2)-b)/(b-1)=(sqrt(5)+1)(sqrt(2)+1)/2
(2-sqrt(2))/(b-1)=((sqrt(5)+1)(sqrt(2)+1)+2)/14
(b-1)=14(2-sqrt(2))/((sqrt(5)+1)(sqrt(2)+1)+2)
b=(14(2-sqrt(2))+(sqrt(5)+1)(sqrt(2)+1)+2)/((sqrt(5)+1)(sqrt(2)+1)+2)
b=(31-13sqrt(2)+sqrt(10)+sqrt(5))/(3+sqrt(5)+sqrt(10)+sqrt(2))~1.84
And with placement 《b》in relationship 1 we have ;
a=(22-b)/(sqrt(2)+1)
a=(22-(1.84))/(sqrt(2)+1)~8.36
R²=(1.84+8.36-8)²/2=2.42
and the equation of the circle
(X-8.36)²+(y-1.84)²=2.42
Don’t use ln().
log_2(512) =9, so use base 2 logs.
could not use power rules to rewrtie as x to x to the 3rd power, then take the 3rd root both sides and end with x to x is a number which will be easier to solve since the other side will be
Thank you, very good method
does anyone know how to do
Sigma,
from n=1 to 2k of (-1)^n+1 * (2n-1)^2
???
Wolfarm alpha can do it and its = 1/2 (-16 (-1)^(2 k) k^2 + (-1)^(2 k) - 1) but I couldn't find the explanation as to why this is the answer I'd really appreciate if someone is educated on the subject and can help me out
You can go by induction or calculate it explicitly.
In the second case you should find out that the sum is equal to:
Sum form j=1 to k of {(4j-3)^2-(4j-1)^2}
Notice that the summatory changed from 2k to k since i compute two terms every time. I write The propg of this statement at the end of the comment.
Then you have (after some calculation)
Sum from j=1 to k of {-2(8j-4) } that leads to -8* sum from j=1 to k of (2j-1)
It is known the the sum of odd sequential numbers is k^2 so you have finally -8k^2
Demonstration of the first fact:
Splitting the sum when n is odd and even you have:
First case: n=2j-1
Sum from n=1 to 2k-1 of {(-1)^(n+1)(2n-1)^2} that leads to
Sum form j=1 to k of {(-1)^(2j) (4j-3)^2
(Notice that -1 is elevated to an even power so is always 1)
Second case: n=2j
Sum from n=2 to 2k of {(-1)^(n+1)(2n-1)^2} that leads to
Sum form j=1 to k of {(-1)^(2j+1) (4j-1)^2
(Notice that -1 is elevated to an odd power so is always -1)
Than you can sum the two summatories with index j and have the result i wrote above.
you can simplify the wolfeam alpha solution since (-1)^2k is 1 and the all solution is -8k^2
@@il_caos_deterministico
Sorry I should have asked for something else,
How do you simplify Sigma from n=1 to any number we'll call it k. Of (-1)^n+1*(2n-1)^2
I'm really sorry Ive wasted your time by not making my question clear..
@@mariothethird5624 it’s not clear still.. i mean, the sum from n=1 to 2k is simply -8k^2, and i have given you how to compute it in the first comment.. you can’t really do something different from what i’ve done (separatong the sum in two parts, reindexing and put them together again)
@@il_caos_deterministico
Well you see, I'm still in high school and Im not even supposed to learn these stuff yet but I'm really curious and want to learn about it.. what I'm trying to say is that I don't even know how to do a general simplification of a sigma going from n=1 to k of any sort of sigma and what I'm asking is do you know what it means to simplify a sigma going from n=something to k (meaning a general number and not a specific number)??
And I dont want to waste your time but if you could just direct me to the right place/tell me what to search on google/youtube to learn about it I would really appreciate it. Btw thank you for even replying in the first place that's very nice of you so thanks.
x=2 answer
512=2^9= 2^3^2
x^3^x=2^3^2
apparently x=2 since the base and the exponent of 3 are the same.
nice problem
but if y=2 , what is value of x in this case ? how can we solve this problem ??
Me seeing 512: It's some power of 2. Maybe try x=2 and wolla it works 😃
Solve this find the equation of circle inscribed in a triangle whose sides are lines L1:x+y=8 L2:2x+y=22 L3:3x+y=22
Note that R = distance from C(x0; y0) to each line
Basically you have to find the equations of two bisectors and intersect them to find the center of the circle, then just use the fact I wrote in the other comment to find the radius R. Then remember that R²=(x0)²+(y0)²-c to find c and complete the equation which is:
x²+y²-2x0x-2y0y+c=0
Find the 3 intersections A (xA,yA), B(xB,yB) , C (xC,yC) of the 3 lines : that is your triangle
Compute the length of sides AB=c, AC=b, BC=a
Recall the incenter I (xI,yI) (the center of inscribed circle in the triangle) is the *centroid of A, B, C with weights a,b,c* --->
a*IA + b*IB + c*IC = 0 ---> projection on oX, oY etc...
Use of Heron formula to have the Area of triangle ABC: Area = sqrt(p*(p-a)*(p-b)*(p-c)) with p = 0.5*(a+b+c)
then the radius r = Area/p -----> equation of inscribed circle : *(x-xI)^2 + (y-yI)^2 = r^2*
x^3^x=2^9
x^3^x=2^3^2
x=2
why take logs base e
take logs base 2
log_2x.3^x=9
instantly leads to x=2
x^{3^x}=512;
x should be > 0;
x^{3^x}=2^9;
if x=2 we have 2^{3^2}=2^9, so x=2 is a solution. Is it the only ? Yes because function x^{3^x} is strictely increasing obviously. Not a very interesting equation.
By inspection, x = 2
Thanks so much because of you i can i think go to mathematical olmypiad
Can there be a negative solution ? Taking logarithms makes sence only if x is positive
.
I think would be more interesting x^2^x=512 without integer solution
x^3^x=2^9
x=2
x=2 by inspection.
That’s it?
x=2 x^3^2 = 2^9 = 2^3^2
And now please do the same But 511 on the right side, instead of 512
x = 2
lembart
2, guess and check u got me
Thnku
Too easy number to solve intuitively x=2 almost trivial
I solve it within 5sec.
Without pen paper
X=2.
👍
1/2 raised to root 3 ~ log 2 . . . just an observation
Wow! They are pretty close