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Very good. Thanks 🙏
Welcome 😊🤩🤩🤩
MIT prof: no aptitude for you!
5(5^√x+√x)=5^2+25^√x+√x-5^2+2√x=2. X=4
Do we need to show that x=4 is the only solution?The left side of the equation is increasing. Therefore it can be equal to 135 only once.
MIT University Entrance Interview: 5^(√x + 1) + 5√x = 135; x =?[5^(√x + 1) + 5√x]/5 = (135)/5, 5^√x + √x = 27; 27 > 5^√x > √x, 3 > √x > 05^√x + √x = 27 = 5^2 + 2 = 5^√4 + √4; x = 4Answer check:5^(√x + 1) + 5√x = 5^(√4 + 1) + 5√4 = 5^(2 + 1) + 5(2) =125 + 10 = 135; ConfirmedFinal answer:x = 4
This is the way to do it. Excellent.
@@moemirza1865Thanks, 🙏
Awesome performance ✅💪
Could you explain me how you derived the 2nd inequality (3 > x^1/2 > 0) ? Thank you.
Why 5²⁷?
見たら一分以内にX=4と解けるよね。答えは135なんだから、√xは整数になる必要がある。つまり、X=1,4,9...になるでしょ。合計して130台になるなら9はありえない。つまり答えは4。ログを使う必要もないし、こんな高校1年レベルの問題に時間をかけてるようじゃ日本の難関大学は受からない。
3:04 Problem solvable "by inspection" right here: m==2; x == 4.
5^{x+x ➖ 1+1 ➖ }+{5+5 ➖} {x+x ➖ }=5{x^2+2}=+{10+x^2}=5^2x^2+10x^2={10x^2+10x^2}=20x^4 2^10x^4 2^2^5x^4 1^1^1x^2^2 x^1^2 (x ➖ 2x+1).
let u=Vx , 1=(27-u)/5^u , / *1/5^(-27) , 5^27=(27-u)/5^(u-27) , 5^27=(27-u)/e^ln5(u-27) , /*ln5 , ln5*5^27=ln5*(27-u)/e^ln5(u-27) , ln5*5^27=ln5*(27-u)*e^(-ln5(u-27)) , ln5*5^27=ln5*(27-u)*e^(ln5(27-u)) , ln5*5^27=ln5*5^2*5^25 , ln5*25*e^(ln5)^25 , ln5*25*e^(ln5*25) , --> , ln5*25=ln5*(27-u) , 25=27-u , u=2 , u=Vx , Vx=2 , solu , x=4 , test , 5^(V(4)+1)+5*V4=5^(2+1)+5*2 , 5^3+10=125+10 , --> 135 , OK ,
Very good. Thanks 🙏
Welcome 😊🤩🤩🤩
MIT prof: no aptitude for you!
5(5^√x+√x)=5^2+2
5^√x+√x-5^2+2
√x=2. X=4
Do we need to show that x=4 is the only solution?
The left side of the equation is increasing. Therefore it can be equal to 135 only once.
MIT University Entrance Interview: 5^(√x + 1) + 5√x = 135; x =?
[5^(√x + 1) + 5√x]/5 = (135)/5, 5^√x + √x = 27; 27 > 5^√x > √x, 3 > √x > 0
5^√x + √x = 27 = 5^2 + 2 = 5^√4 + √4; x = 4
Answer check:
5^(√x + 1) + 5√x = 5^(√4 + 1) + 5√4 = 5^(2 + 1) + 5(2) =125 + 10 = 135; Confirmed
Final answer:
x = 4
This is the way to do it. Excellent.
@@moemirza1865
Thanks, 🙏
Awesome performance ✅💪
Could you explain me how you derived the 2nd inequality (3 > x^1/2 > 0) ? Thank you.
Why 5²⁷?
見たら一分以内にX=4と解けるよね。答えは135なんだから、√xは整数になる必要がある。つまり、X=1,4,9...になるでしょ。合計して130台になるなら9はありえない。つまり答えは4。ログを使う必要もないし、こんな高校1年レベルの問題に時間をかけてるようじゃ日本の難関大学は受からない。
3:04 Problem solvable "by inspection" right here: m==2; x == 4.
5^{x+x ➖ 1+1 ➖ }+{5+5 ➖} {x+x ➖ }=5{x^2+2}=+{10+x^2}=5^2x^2+10x^2={10x^2+10x^2}=20x^4 2^10x^4 2^2^5x^4 1^1^1x^2^2 x^1^2 (x ➖ 2x+1).
let u=Vx , 1=(27-u)/5^u , / *1/5^(-27) , 5^27=(27-u)/5^(u-27) , 5^27=(27-u)/e^ln5(u-27) , /*ln5 ,
ln5*5^27=ln5*(27-u)/e^ln5(u-27) , ln5*5^27=ln5*(27-u)*e^(-ln5(u-27)) , ln5*5^27=ln5*(27-u)*e^(ln5(27-u)) ,
ln5*5^27=ln5*5^2*5^25 , ln5*25*e^(ln5)^25 , ln5*25*e^(ln5*25) , --> , ln5*25=ln5*(27-u) , 25=27-u , u=2 , u=Vx ,
Vx=2 , solu , x=4 , test , 5^(V(4)+1)+5*V4=5^(2+1)+5*2 , 5^3+10=125+10 , --> 135 , OK ,