Solving a 'Stanford' University entrance exam

Solving a 'Stanford' University entrance exam | m=?

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Күн бұрын

Пікірлер: 18

@Zxcdd851 22 сағат бұрын

The solution is correct but not elegant, I would rather put it this way : 9^(4^m)=4^(9^m) exp(ln(9).4^m)=exp(ln(4).9^m) 2.ln(3).exp(ln(4).m)=2.ln(2).exp(ln(9).m) ln(3).exp(ln(4).m)=ln(2).exp(ln(9).m) ln(ln3))+m.ln(4)=ln(ln2))+m.ln(9) ln(ln3))+m.2.ln(2)=ln(ln2))+m.2.ln(3) 2.m.(ln(3)-ln(2))=ln(ln(3))-ln(ln(2)) m = 1/2.{ln(ln(3)-ln(ln2))}/{ln(3)-ln(2)}

@AsadInternationalAcademy 32 минут бұрын

nice explanation

@AsadInternationalAcademy 32 минут бұрын

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@اسماعیلخسروی-خ6ظ 2 күн бұрын

❤❤❤

@AsadInternationalAcademy 2 күн бұрын

thank you so much

@AsadInternationalAcademy 2 күн бұрын

@andresrebolledobanquet1924 2 сағат бұрын

36m = 36m

@AsadInternationalAcademy 38 минут бұрын

@ConradoPeter-hl5ij 23 сағат бұрын

I saw a guy commenting on yours, so I decided to try to help you. The accent in your pronunciation is very different from that of the native English speaker. You are pronouncing with a very strong high pitch sound in syllables that should be softer. This creates tension in the listener. Try speaking in a slightly lower tone next time.😊 I hope helped

@AsadInternationalAcademy 31 минут бұрын

@ConradoPeter-hl5ij Күн бұрын

9^4^n=4^9^n (3²)^(2²)ⁿ=(2²)^(3²)ⁿ (3²)^(2²ⁿ)=(2²)^(3²ⁿ) [3^(2²ⁿ)]²=[2^(3²ⁿ)]² |3^(2²ⁿ)| = |2^(3²ⁿ)| => 3^(2²ⁿ) = 2^(3²ⁿ) or 3^(2²ⁿ) = -2^(3²ⁿ); but for n € R this is impossible. => log[3^(2²ⁿ)]=log[2^(3²ⁿ)] => (2²ⁿ)log3=(3²ⁿ)log2 => (2²ⁿ)/(3²ⁿ)=log2/log3 => (2/3)²ⁿ=log2/log3 => (2/3)²ⁿ≈(0,301)/(0,477) => (2/3)²ⁿ≈301/477 => (2/3)²ⁿ≈300/477; an acepptable error => (2/3)²ⁿ≈100/159 => (2/3)²ⁿ≈100/160 => (2/3)²ⁿ≈10/16 => (2/3)ⁿ≈(10/16)½ => (2/3)ⁿ≈(√10)/4 => log[(2/3)ⁿ]≈log[(√10)/4] => n[log(2/3)]≈log(√10) - log4 => n[log2 - log3] ≈ log(10½) - log2² =>n[0,301-0,477]≈(1/2)log10-2log2 => n(-0,176)≈(1/2)(1) - 2(0,301) => n(-0,176) ≈ (1/2) - (0,602) => n(-0,176) ≈ 0,5 - 0,602 => n(0,176) ≈ 0,602 - 0,5 => n(0,176) ≈ 0,102 => n ≈ (0,102)/(0,176) => n ≈ 102/176 => n ≈ 51/88 => n ≈ (55-4)/88 => n ≈ (55/88) - (4/88) => n ≈ (5/8) - (1/22) => n ≈ (5/2³) - (1/22) => n ≈ (5⁴/10³) -(1/22); 5⁴=25²=625 => n ≈ (0,625) - (1/22) => n ≈ (0,625) - (1/2)(1/11) => n ≈ (0,625) - (1/2)(9/99) => n ≈ (0,625) - (1/2)(0,0909090...) => n ≈ (0,625) - (0,0454545...) => n ≈ 0,580 - (0,000454545...) => n ≈ 0,579545454... => n ≈ 0,5795 (300159) => n ≈ 0,579 (I will consider that way, cuz I made modifications before) a=4ⁿ≈2,2314786446 b=9ⁿ≈3,5686701889 p=9^a≈134,7012982852 q=4^b≈140,7840786512 (a good approximation) In fact, n ≈ 0,5679412819 э ≈ 0,579 - 0,5679412819 э ≈ 0,0129412819 э ≈ 1,3×10-³ (an acceptable error) 9^4^(0,5679412819) ≈ 125,019025 4^9^(0,5679412819) ≈ 125,019024

@AsadInternationalAcademy 30 минут бұрын