Japanese || Math Olympiad || Can you solve this ? || Square Root Simplification😊

[@RyanLewis-Johnson-wq6xs]

Sqrt[5^xSqrt[25^xSqrt[125^x]]]=125 x=24/11 x=2.18 recurring

[@souadnait6421]

Thanks 😊👍

[@superacademy247]

You're welcome!

[@AbbasGuclu-jn3po]

x=24/11

[@key_board_x]

√[5^(x) * √{25^(x) * √[125^(x)]}] = 125 → recall: √a = a^(1/2)
√[5^(x) * √{25^(x) * 125^(x)]^(1/2)}] = 125
√[5^(x) * √{25^(x) * 125^(x/2)}] = 125
√[5^(x) * {25^(x) * 125^(x/2)}^(1/2)] = 125
√[5^(x) * 25^(x/2) * 125^(x/4)] = 125
[5^(x) * 25^(x/2) * 125^(x/4)]^(1/2) = 125
5^(x/2) * 25^(x/4) * 125^(x/8) = 125 → you know that: 25 = 5^(2) and you know that: 125 = 5^(3)
5^(x/2) * [5^(2)]^(x/4) * [5^(3)]^(x/8) = 5^(3)
5^(x/2) * 5^(2x/4) * 5^(3x/8) = 5^(3)
5^(x/2) * 5^(x/2) * 5^(3x/8) = 5^(3)
5^[(x/2) + (x/2) + (3x/8)] = 5^(3)
(x/2) + (x/2) + (3x/8) = 3
x + (3x/8) = 3
8x + 3x = 24
11x = 24
x = 24/11

[@superacademy247]

Stunning 👌 method. Thanks

[@prollysine]

let u=5^x , sqrt(u*sqrt(u^2*sqrt(u^3)))=125 , ₁ , u^(3*1/2)=u^(3/2) , ₂ , u^(3/2+4/2)=u^(7/2) , ₃ , u^(7/2*1/2)=u^(7/4) ,
₄ , u^(7/4+4/4)=u^(11/4) , ₅ , u^(11/4*1/2)=u^(11/8) , ₆ , recall u=5^x , --> (5^x)^(11/8) , 5^((11/8)x)=125 , 125=5^3 ,
(5^x)^(11/8)=5^3 , ()^(8/11) , ₇ , (5^x)(11/8*8/11)=(5^3)^(8/11) , x=3*(8/11) , ₈ , x=24/11 ,
solu. , x=24/11 , test , 5^(3x)=5^(72/11) , 5^(72/11)*(1/2))=5^(72/22) --> 5^(36/11) , 5^((36/11)+(2*24/11))=5^(84/11) ,
5^((84/11)*(1/2))=5^(84/22) , --> 5^(42/11) , 5^((42/11)+24/11))=5^(66/11) , 5^((66/11)*(1/2))=5^(66/22) ,
5^(66/22)=5^3 , 125=5^3 , same , OK ,

[@superacademy247]

Great 👌 solution

[@prollysine]

@@superacademy247 Thank you very much, I would like to understand math better..

[@souadnait6421]

Thanks 😊👍

[@superacademy247]

Welcome 😊

[@souadnait6421]

Thanks 😊👍

[@superacademy247]

No problem 😊