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This is so cool Professor!
This one is quite unique in my opinion
Well explained sir
V nice explaining the ruleof indicesv clearly
Thnku
Wow this one was too easy
5th root 10
let y=x^5, get x = y^(1/5)
-> (y^(1/5)^y = 10^2
-> y* lg(y^(1/5)) = 2
-> y * lg(y) = 10
-> y^y = 10^10
-> y = 10
-> x = y^(1/5) = 10^(1/5)
Nice video sir
Nice method
i made the substitution u=x^5 so
u^1/5u=100
u^u=100^5=10^10
u=10
x=5th root of 10
There is an issue:
a^b^c = a^(b^c) or a^b^c = (a^b)^c,
in general case a^(b^c) is not equal to (a^b)^c.
So, I think we should clarify this issue before solve the problem……
Nice and easy question
a^a=b^b———>a=b ,if and b ,a>=1 or if 0.368<a,b<=1
Thanks for video. Good luck sir!!!!!!!
nice.
remember my JEE advance preparetion time.
nostralgic
https://youtu.be/1Jp6IcxM8j4 this is new math channel
Outstanding explain
a^a =b^b implies that a=b? Is that always true? A graph of y=x^x curves back on itself near zero so has repeat values so I would say this is not generally true. Anyone?
Siuuuuu, thanks for shearing
I have learn another word!! Xd
I used Lambert W function to solve this: x^5 ×lnx=ln100 —》x^5 × ln(x^5)=ln( 10^10)—》W(ln(x^5) × e^(ln(x^5)))=W(10ln10)—》ln10=ln(x^5)—》x=10^0.2
I got as far as X*LOG(X)=2/5 before the crystal ball went dark. I had no clue. Very clever solution. Thanks!
very well done, thanks for sharing this Math Olympiad question
I like multi-decker exponents. It looks so nice<)) Thank you so much, Mr PreMath! Stay blessed and all the best to you!
Beautiful solution
Excellent sir
Thanks sir.love from Bangladesh
Herr professor, I saw your solution and I didn't understand it, as I believe that the value of X = 1.7124
How I made: X^X = Y
2) Y^5 = 100
Y = 2,5118
2,5118 = X^X
X = 1,7124
where did I go wrong?
Well, I got that completely wrong – most interesting and I thank you!
amazing …https://youtu.be/77W9zhTBuUM
x^x^5=100 add 100 to both sides
x^x^5=10^2 convert 100 to exponent form
(x^x^5)^5= (10^2)^5 raised both sides to the power of 5
x^5^x^5=10^10 now there is a pattern, x^5 's base matches 10 (base), and x^5's exponent matches 10 (exponent) hence
x^5=10
x= 10^1/5 or x=10^0.2
I wish the video had discussed the secondary root of 100, so that when 100 was rewritten as 10^2 we also looked at (-10)^2
✅
Manipulation at its best!
x=10^1/5 or 10^0.2 or 1.5849
fifth root of 10
Very well done. I like your technique. However, I got the same answer you did by putting the equation into the "Lambert form":
ln x^(x^5) = ln 100
x^5 [ln x] = 4.605
e^ [ln (x^5)] * ln x = 4.605
e^ [5 ln (x)] * lnx = 4.605
5 ( e^ [5 ln (x)] * lnx) = 5* 4.605
( e^ [5 ln (x)] * 5 lnx) = 23.025
Lambert both sides and get. . .
5 lnx = productlog (0, 23.025) from Wolfram Alpha
5 ln x = 2.30256
ln x = 0.460512
e^(ln x) = e^0.460512 is approx. 1.585 which also equals approx. 10^1/5
x = 1.585.
Nice to get the same answer with two different techniques.
Only if you test the solution
x= 10^(1/5)=1.58489319
then you get with
(x^x) = 2.07481146
and
(x^x)^5= 38.44972 and NOT 100!
Where am I wrong?
@PreMath I am really proud to be on your channel, I am an electrical engineer,47 yrs old & I have liked Math since I was a child, when I watch your videos & try solving before seeing the solution, it's a great pleasure, so many thanks for you
Can u solve it with log?
There is a easier way to do it
X^5=t
So X= t^1/5
X^X^5=100
Log base 10 on both sides
X^5logX=2log10
t/5logt=2
tlogt=10
t^t=10^10( because of log base 10)
t=10
So X= t^1/5=10^1/5
If y is substituted for x^5 the equation becomes y^(y/5) = 10^2 and the solution is obvious.
Like
ha ha ha….
Shouldnt this problem have 5 roots?
Let X =2y
好
好，奖 1 万元
But what is 10 to the 1 fifth?
By inspection, x==1 is too small. X==2 gives 2^32 which is too big. So 1 < x < 2. Trying the mean value x == 3/2 looked close. I could not be bothered doing any more than that.