Nice Math Olympiad Radical Simplification | How to find "m" ?

[@charlesrowe310]

it might be faster to convert the square roots to exponents & distribute. Then add them up. You are quickly left with m^(7/8) = 2^7 fraction rule to rewrite will get you 8th root of m = 2 then you can be left with m = 2^8 Either way, it was an excellent video. thank you for it!

[@prollysine]

128= 2^7 , the m power , 1/2 , 3/2 , 3/4 , 7/4 , 7/8 , m^(7/8)=2^7 , / ()^(8/7) , m=(2^7)^(8/7) , m=2^8 , m=256 ,

[@MathBeast.channel-l9i]

Right 👍

[@prollysine]

@@MathBeast.channel-l9i Thanks!

[@AvikDey-z6y]

√(m) √(m) √(m) =128 Or m√(m) √(m) =128*128 Or m^2*m*m√m=128^4 Or m^2×m^2×√m=128^4 Or( m^4) ^2×(√m) ^2= 128^4 Or m^9=2*2*2*2*2*2*2*2*2 =2^9 Now m=2

[@MathBeast.channel-l9i]

@@AvikDey-z6y Can you please check it out to verify?

[@RealQinnMalloryu4]

2^7 2^7^1 2^1^1 2^1 (m ➖ 2m+1).

[@arekkrolak6320]

Is this video a joke? Taking 7th root gives you 7 different complex solutions on the left side and 7 on the right side of the equation. So the guy arbitrarily chose 1 if them ignoring remaining 48 and claims it a solution :)

[@ByzasTT]

How Oh and Can I ask one more question? In ГmГmГm=128 Ismt it M^1/2*m^1/2*m^1/2=128? And by indices law multiplication is addition so (M^3/2)=128 When you square it'll be M^3=128 Why did only one root get erased Am I missing something?

[@jamesharmon4994]

m^(7/8) = 2^7 Taking 7th root both sides m^(1/8) = 2 Taking both sides to the 8th power m = 2^8 aka 256

[@MathBeast.channel-l9i]

@@jamesharmon4994 Nice Approach 😊

[@nasrullahhusnan2289]

LHS=m^(½+¼+⅛) =m^⅞ 128=2⁷ Hence m^⅞=2⁷ --> m^⅛=2 m=2⁸ --> m=256

[@MaheshKhatri-in6bc]

m=256