Solve in one Minute | Learn how to solve exponential equation quickly | Math Olympiad Training -

Solve in one Minute | Learn how to solve exponential equation quickly | Math Olympiad Training

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Can you solve the given exponential equation for x in one minute?

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Solve in one Minute | Learn how to solve exponential equation quickly | Math Olympiad Training

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  1. V nice explaining the ruleof indicesv clearly

  2. 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)

  3. i made the substitution u=x^5 so
    x=5th root of 10

  4. 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……

  5. a^a=b^b———>a=b ,if and b ,a>=1 or if 0.368<a,b<=1

  6. nice.

    remember my JEE advance preparetion time.


  7. 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?

  8. Siuuuuu, thanks for shearing
    I have learn another word!! Xd

  9. 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

  10. I got as far as X*LOG(X)=2/5 before the crystal ball went dark. I had no clue. Very clever solution. Thanks!

  11. very well done, thanks for sharing this Math Olympiad question

  12. I like multi-decker exponents. It looks so nice<)) Thank you so much, Mr PreMath! Stay blessed and all the best to you!

  13. 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?

  14. Well, I got that completely wrong – most interesting and I thank you!

  15. 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= 10^1/5 or x=10^0.2

  16. 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

  17. 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.

  18. Only if you test the solution
    x= 10^(1/5)=1.58489319

    then you get with
    (x^x) = 2.07481146
    (x^x)^5= 38.44972 and NOT 100!

    Where am I wrong?

  19. ​ @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

  20. There is a easier way to do it
    So X= t^1/5
    Log base 10 on both sides
    t^t=10^10( because of log base 10)
    So X= t^1/5=10^1/5

  21. If y is substituted for x^5 the equation becomes y^(y/5) = 10^2 and the solution is obvious.

  22. 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.

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