Is it really Oxford test ? Easily can be resolved in brain...I would use 10, not 4, for the basis...😂

7th grade students will do that: k*k*k - k*k = 100 >>> k^2 (k-1) = 100 >>> (k-1) =100 / k^2 👉 (k-1) is an integer >>> 100 divisible by (k^2) >>> (k^2) = 2 4 5 10 20 25 50 😁 👉 (k^2) is the square number >>> k^2 = 25 >>> k = 5 😁 Very simply 👍

Once I got 2^16*15, I turned that into 2^10 (1024) times 2^5*3 (32*3, 96 or 100-4) times 2*5 (10) and all that multiplication became a fairly easy subtraction problem. (102,400-4096)*10 98,304*10

There's no domain restriction on p, q, and r. 2^p + 2^q + 2^r = 42 , p < q < r Let p = 0 and q = 1 . 2⁰ + 2¹ + 2ʳ = 42 2ʳ = 39 r = log₂(39) One of an infinite number of solutions: p = 0 , q = 1 , r = log(39)/log(2) ≈ 5.2854022188622... Alternate solution: 42 = 2Ah = 101010b = 2¹ + 2³ + 2⁵ p = 1 , q = 3 , r = 5

3051.873195268761708828

X= log 5/{log (3+*5)- log 2} or X=log 5/{ log (3-*5)-log 2}..... May be *= read as to the power ^=read as square root As per question 25^(1/*x)+125^(1/*x)=625^(1/*x) 5^(2/*x)+5^(3/*x)=5^(4/*x) Let 5^(1/*x)=R So, R^2+R^3=R^4 R^2(1+R)=R^4 1+R=R^4/R^2 1+R=R^2 R^2-R-1=0 Here a=1,b=-1, c=-1 D=b^2- 4ac =(-1)^2-(4×1×-1) =1+4=5 *D=*5 R=(-b±*D)/2a ={-(-1)±*5}/(2×1) =(1±*5)/2 R^2=(1+5+2*.*5)/4 or (1+5-2.*5)/4 R^2=(6+2.*5)/4 or (6-2.*5)/4 =2(3+*5)/4 or 2(3-*5)/4 =(3+*5)/2 or (3-*5)/2 Again R^2={5^(1/*x)}^2 =[5^{(1/x)^(1/2)}^2 =5^(1/x) So, 5^(1/x)=(3+*5)/2 Take the log log {5^(1/x)}=log{(3+*5)/2} (1/x). log 5 =log(3+*5) - log 2 (1/x)={log(3+*5)-log2 }/log 5 X=log 5/{log(3+*5) -log2 } Similarly Taking the log we'll get X=log 5/{log(3-*5) - log2 }

It took him nearly ten minutes, to do what took me a minute by simply multiplying 4 in my head followed by a simple subtraction.

1/Vx=log((1+V5)/2)/log5 , x=(log5/log((1+V5)/2))^2 , x=~ 11.186027 ,

k³ - k² - 100 = 0 We can see k = 5, so (k - 5) is a factor: (k - 5)*(k² + 4*k + 20) = 0 k = 5, -2 ± i*4

You are the best example of how an easy thing can be made tough 👍🏻👍🏻

25^(1/Sqrt[x])+125^(1/Sqrt[x])=625^(1/Sqrt[x]) X= (Log[(1+Sqrt[5])/2,5])^2=Log[(((5/10)+(5/10)Sqrt[5])^Ln(5/10+(5/10)Sqrt[5])),5^Ln(5)/e^(11Ln(((5Sqrt[5]-5)/10)^Ln(5Sqrt[5]-5/10)))]+11 X≈ 11.186027253724131066250820416140739258619523113454702829972714855178241686535009282270443557381505996043133508614761773275855980701943550865892526959700048076260917556733405660435089930337225134775515509138269419891185814842006109671976106796298195009431867908378952252189600033383759518660176990768891668889308072338702315567190672239152772727354303149818020180367713429260553254581683753669789749733257276414767917029912030543558986280818312505378724541892518439024767898898479375675677165035554296619979829119612875522127395491671734596612676078381494416296242335436691273066315837192770699551063226948259203459255078736122522129817909929256665000759505736113449334921216422898991728031292021634043569113728094885129030980608727750774392702158454055771440018620078548804770139971830999911593398247426010227753599027218742226606176854329741165187092973034836760262675231917038724076372950626017444884038808277580909312091840208986338640204452944030728369112502315659925869905347080022261809589100814388846583380503063509387785577049004687272634700159893323310573901900280140054717777960555850540077422827513881981603068025169396094793346597422991332722161681284508129435684124895408627722954140425148378449127157487300974229864276752543927163263007124465208366769371571498374302867462918320818946341854710942173041832873092283140142825989365792899031477849357546930422889231153704266812920171340910620534277697613729353692489131118326208733202493404994480178211396591126373600434138622909128914472294013521343189714831813806812622613752993627010270887125117337332202848533465865864868729473507545122677392091636190053907910303526656885049700858352837539295024490828463654576374309388980742862265174362630345271464921238826405372711766314393466131541046394820136514601334332986871519289630573122939740192759692898817559307248163082374462570306825893950969762382565805835652955815191057792703967195216726836342422497070177326263411128502342828077773255864278347610853355927590241437608553983506143531218737091323017863806696583071647432201004314726208301427331899585073527698371...

A Douglas Adams "Hitch Hikers" guide to the Galaxy reference. The computer gives the answer "42" to the question "life, the universe, and everything". Given computers today use binary, the question assumes binary is still used, and asks the question, How would the computer represent 42 ?

Oxford admission to what ? To the "course of equity and verieties " 😂😂😂

Question: At 6:40 you say +2+27 is 25. Isn't it 29? Thanks for helping.

x-y = 2, x = 2 + y 2^(2+y) - 2 ^y = 4 2^y * (2^2 - 1) = 4 2^y = 4/3 y = (log 4-log3) / log2 = 0.4150 X = 2.4150

method.. assume m is an integer.. find the value of m such that 5^m is the lowest value greater than the answer. powers of 5 are 5,25,125,625… so try m=4. 5^4=625 3^4=81 subtracting gives 544 so m=4 is the answer. 1:57 method given is just a different way of getting this by assuming that (aa-bb)(aa+bb) will give an answer which it does (25-9)(25+9)=16*34=544

4=2^X-2^Y=2^Y（2^（X-Y）-1）=2^Y（2^2-1）=3*2^Y，2^Y=4/3， Y=LOG（4/3）/LOG（2）=2-LOG（3）/LOG（2），X=4-LOG（3）/LOG (2)

Great video, Super Academy. Looking forward to seeing your next upload from you. I smashed the thumbs up button on your content. Keep up the fantastic work! Your breakdown of the quadratic equations was really clear. Could you elaborate on how you approach factoring in more complex scenarios?

It's a factoring quiz & we may solve it from RHS. 42=2*3*7=2*(2^2-1)*(2^3-1) =(2^3-2)*(2^3-1) =2^6-3*2^3+2 =2^6-(2^2-1)*2^3+2 =2^6-2^5+2^3+2 =2^5(2-1)+2^3+2 =2^5+2^3+2

I don't think you needed to use the power rule for logs as it’s implied from canceling the base with the log of the same base leaving with whatever is in the exponent on one side of the equation.

Nice 👍

42=32+8+2 2¹+2³+2⁵ (2^p)=2¹ (2^q)=2³ (2^r)=2⁵ p=1, q=3,r=5

1/x=y y+rt(1+y)=5 rt(1+y)=5-y y<5 1+y=25-10y+y^2 y^2-11y+24=0 (y-3)(y-8)=0 y<5. So, y=3 x=1/3

(1/x)+√[(x+1)/x]=5 (1/x)+√[1+(1/x)]=5 Let y=(1/x) y+√(1+y)=5 5-y=√(y+1) y²-10y+25=y+1 y²-11y+24=0 (y-8)(y-3)=0 y-8=0 y=8 (1/x)=8 8x=1 x=⅛ ∅ (1/⅛)+√[(⅛+1)/⅛]=5 8+√9=5 8+3=5 11≠5 false y-3=0 y=3 (1/x)=3 x=⅓ ❤ (1/⅓)+√[(⅓+1)/⅓]=5 3+√4=5 3+2=5 5=5 ✓

Bạn có mỏi tay lúc viết những số như:10,000,000,000,000. Tôi đề nghị thay thế bằng 13 (13 số 10) góc trên trái và góc trên phải của 13 là 10 nhỏ. (10) 13 (10) Ví dụ khác: 1,111,111,111 là (1) 10 (1). Đây mới là ý tưởng giảm thiểu thời giờ và tăng sự chính xác về Số quá dài. 😂😂

Solution by insight 2^p+2^q+2^r=42 Among 2,4,8,16,32 2+8+32=42 The only possible combination. p=1, q=3,r=5

Waste time

(1/x) + √[(x + 1)/x] = 5 → where: x ≠ 0 (1/x) + √[1 + (1/x)] = 5 → let: a = 1/x a + √(1 + a) = 5 → where: a > - 1 √(1 + a) = 5 - a [√(1 + a)]² = (5 - a)² 1 + a = 25 - 10a + a² a² - 11a + 24 = 0 Δ = (- 11)² - (4 * 24) = 121 - 96 = 25 a = (11 ± 5)/2 a = 8 → x = 1/8 a = 3 → x = 1/3 Frirst case: x = 1/8 → let's check = (1/x) + √[(x + 1)/x] → when: x = 1/8 = (1/{1/8}) + √[({1/8} + 1)/{1/8}] = 8 + √[({1/8} + {8/8})/{1/8}] = 8 + √[(9/8)/{1/8}] = 8 + √9 = 8 + 3 = 11 → x = 1/8 is not a solution Second case: x = 1/3 → let's check = (1/x) + √[(x + 1)/x] → when: x = 1/3 = (1/{1/3}) + √[({1/3} + 1)/{1/3}] = 3 + √[({1/3} + {3/3})/{1/8}] = 3 + √[(4/3)/{1/3}] = 3 + √4 = 3 + 2 = 5 → x = 1/3 is a solution

The primary school pencil-and-paper square root method is much quicker and always gives the result.

A can't understand the step on 4:28, explain me please!

Just write it as a binary number. 32=10101 then convert back to base 10. 10101=32 + 8 + 2 = 2^5 + 2^3+2^1so P is 5, Q is 3 and r =1 (only works when P,Q&R are whole numbers)

You didn’t simplify anything, just a not easy way to calculate the numbers. While the original problem is also kind of stupid!

This solution is getting difficult for this easy problem.....))

No one is going to ask this question!!Its about the SAT or ACT.

a very unique case who's solution has NO UNIVERSAL APPLICATION...

(a-13)(a^3+13a^2-197a-2562)=0 , a=13 , b=a^2-183 , b=-14 , a^3+13a^2-197a-2562=0 , (a+14)(a^2-a-183)=0 , a= -14 , b=13 , sulu , (a,b) = (13 , -14) or (-14 , 13) , / a^2-a-183=0 , a= (1+/-V(1+4*183))/2 , a=(1+V733)/2 , b=(1+V733)/2 , ---- a=(1-V733)/2 , b=(1-V733)/2 , a=b , not a solu , / ,

answer in two steps 1/8^1/8 = 2^(-3)^2^(-3)= (2)^-3*1/8= (2)^-3/8

The first and last terms reduce to 1, and the middle term reduces to -1. -1 ** 7= -1.

This is a question of finding the binary representation of 42. Take r>q>p and define r to be the largest number for which 2^r ≤ 42. Subtract and repeat the procedure for the remainder (here 10) until the remainder is zero. Here we find 42 = 32 + 8 + 2 = 2^5 + 2^3 + 2^1. I.e, r, q, p = 5, 3, 1 or a permutation thereof.

Another way si to try to think with binary Numbers. The solution has to be written as 1 + x1 x 2 + x2 x 4 + x3 x 8 +…. If you make the calculus you can see that the first xi (up to 24) will be zéro and the next ones will be 1 up to i equal 48. In the end 1 + 2^25 + 2^26 +••• + 2^48 equal 1 + 2^49 - 2^25

Square root of 9 =minus -3 and +3 isnt it? So if we take X=equal to minus 3 ,it verifies the equation

If you write 42 in binary, you get 101010 and immediately the unique solution. I think that was what the test expected.

Harvard University Admission Interview Tricks: 1/x + √[(x + 1)/x] = 5; x =? 1 > x > 0; 1/x + √[(x + 1)/x] = 1/x + 1 + √[(x + 1)/x] = 5 + 1 (x + 1)/x + √[(x + 1)/x] = 6, Let: y = √[(x + 1)/x]; y > 0 (x + 1)/x + √[(x + 1)/x] = y² + y = 6, y² + y - 6 (y - 2)(y + 3) = 0, y > 0, y + 3 > 0 y - 2 = 0, y = 2 = √[(x + 1)/x], (x + 1)/x = 2² = 4, 4x = x + 1; x = 1/3 Answer check: x = 1/3, 1/x = 3: 1/x + √[(x + 1)/x] = 3 + √[3(1/3 + 1)] = 3 + 2 = 5; Confirmed Final answer: x = 1/3

you didnt need the W function. in the middle of that you found 2^5+5=37 and since x and 2^x are monotonic increasing functions and thrhs is constant. there is only one solution so x=5

t=1/x...t+√(1+t)=5...1+t=25+t^2-10t...t^2-11t+24=0...t=(11+5)/2=8...t=(11-5)/2=3..x=1/8,1/3..1/8ko

Minus one half, by inspection. The turgid explanation goes off the rails at 4:28, but a pile of obscure errors around 6:44 result in the correct answer by accident. There is no planet in this Universe where two is the natural log of four, but it's close enough to get there.

How long was the time for this question?

This is all incorrect. The eighth root of anything is the square root of the square root of the square root. Here that comes down to 1/(2^.5) or half of the root of two, i.e. roughly 0.71.

My solution is way easier (it took less than a minute): 1) I rewrote the equation as 1/x+sqrt(1/x+1)=5. 2) Then, I added 1 to both sides of the equation 1/x+1+sqrt(1/x+1)=6. 3) Then, by simply substituting t=sqrt(1/x+1), we get a simple quadratic equation of t^2+t-6=0. The solution of which are -3 and 2. 4) When we plug in the value of sqrt(1/x+1)=-3 has no real solutions ( by the way, if it did not violate the laws of mathematics, the answer choice of 1/8 would be correct). 5) sqrt(1/x+1)=2 1/x+1=4 x=1/3

x^2 + 3x = 10000099998 x^2 + 3x = 10000000000 + 99998 x^2 + 3x = 100000*100000 + 99998 let y = 99999 x^2 + 3x = (y+1)*(y+1) + (y-1) x^2 + 3x = y^2 + 2y + 1 + y -1 x^2 + 3x = y^2 + 3y x^2 + 3x - y^2 - 3y = 0 (x^2 - y^2) + 3(x-y) = 0 (x-y)*(x+y) + 3*(x-y) = 0 (x-y)*(x+y+3) = 0 either (x-y)=0 ==> x = y = 99999 or (x+y+3) = 0 ==> x = -(y+3) = -(99999+3) = -100002