99% of Students Fail Oxford University Entrance Exam Because of THIS

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  • Опубліковано 17 січ 2025

КОМЕНТАРІ • 23

  • @joeviolet4185
    @joeviolet4185 3 місяці тому +4

    If I look at the given equation for about 30 seconds I see the solution and do not need any substitutions or Lambert w functions and 3 sheets of paper filled with calculations.

  • @manuelgonzales2570
    @manuelgonzales2570 3 місяці тому +3

    Excellent exercise! Thank you!

  • @X00000370
    @X00000370 3 місяці тому +1

    Another good start to the morning, more math fun!

  • @michaeledwards2251
    @michaeledwards2251 3 місяці тому

    For 27^x + x = 0
    Inspection reveals
    1. (a) if x = 0, 1 + 0 > 0, If x > 0, 27^x + x > 0, by elimination x is negative.
    (b) if x = -1, 1/27 -1 < 0, x > -1.
    x is a negative fraction with an absolute value less than 1.
    2. Factoring 27 gives 3^3 or 3*3*3.
    3. (3^3)^x = - x by subtracting x from both sides of 27^x + x = 0.
    4. x is a negative fraction giving ( 1/((3)^(3*-x))) = -x.
    5. The multiple of (3*--x) is an integer, an irrational value would make 27^x and x, irrational.
    6. The only value of x giving an integer value of (3* -x) is -1/3. ( - - is + )
    7. Insert x = -1/3 into 27^x + x = 0 gives (27^(-1/3) = 1/3 and x = (-1/3)) or (1/3) + (-1/3) = 0
    Giving an answer of x = -1/3 : no evidence of solution by inspection among Oxford entrants.

  • @phuocvlog
    @phuocvlog 3 місяці тому +1

    u^u=3^3 => u =3 ìf you can prove that f(x)=x^x is monotonic function.

  • @vortexlegend101
    @vortexlegend101 3 місяці тому

    27 is a perfect cube, so it makes sense to take the cube root (x =1/3), but since 27^x is always positive, x must be negative, so I tried x = -1/3, which worked out as a solution.
    While your method is much more rigorous, in an entrance exam it is more useful to be good at spotting answers quickly and leaving more time to work on harder questions. I do think the methods in the top comments are more reliable than my guesses though, so I’d use them!

  • @walterwen2975
    @walterwen2975 3 місяці тому +2

    Oxford University Entrance Exam: 27ˣ + x = 0; x =?
    x ≠ 0; x = - 27ˣ, x¹⸍ˣ = (- 27ˣ)¹⸍ˣ = - 27 = (- 3)³
    x¹⸍ˣ = (- 3)⁽⁻¹⁾⁽⁻³⁾ = (- 3⁻¹)⁽⁻³⁾ = (- 1/3)¹⸍⁽⁻¹⸍³⁾; x = - 1/3
    Answer check:
    x = - 1/3: 27ˣ + x = 3³⁽⁻¹⸍³⁾ + (- 1/3) = 3⁻¹ - 3⁻¹ = 0; Confirmed
    Final answer:
    x = - 1/3

  • @raghvendrasingh1289
    @raghvendrasingh1289 3 місяці тому

    27^x = -x
    (- x )^ (1/x) = 27
    (-1/x)^ (-1/x) = 3^ 3
    -1/x= 3
    x= -1/3
    Now derivate of 27^x+x is
    27^x log 27 +1 which is always positive
    Hence function is increasing and only solution is x= -1/3

  • @richardcarnegie777
    @richardcarnegie777 3 місяці тому

    It took 4 simple lines of algebra to get the W-function solution.

  • @nigellbutlerrr2638
    @nigellbutlerrr2638 3 місяці тому +1

    Engineering Mathematics is Interesting.
    A lot of pure maths appears to be Solomon's Vanity.😮😮. But people are blind 🦮.

  • @CharlesChen-el4ot
    @CharlesChen-el4ot 3 місяці тому +1

    X=-1/3
    27^x = 1/27^1/3
    =1/3
    27^x + x = 0

  • @9허공
    @9허공 3 місяці тому

    you must show that there is only one solution.
    27^x = -x , LHS is strictly increasing , RHS is strictly decreasing
    => there is only one solution => x = -1/3.

  • @saurabhsrivastava4140
    @saurabhsrivastava4140 3 місяці тому

    27^x=-x
    3^3x=-x
    LHS is positive & RHS is negative
    Put value of x in negative
    Check Put x=(-1/3)
    3*(-1/3)=-1
    We can see that, LHS= RHS
    So, x=-1/3 is answer

  • @SinCosTan1by0
    @SinCosTan1by0 3 місяці тому

    -1/3

  • @alecrego4879
    @alecrego4879 3 місяці тому

    Took me like 5 seconds

  • @benjaminkarazi968
    @benjaminkarazi968 3 місяці тому

    Hello,
    Oxford University Fails Entrance Exam Because
    -1÷3=-3.33333333333e⁻¹ limit ➟ ∞ digits
    ➪ 27ˣ + x = 27⁻³.³³³³³³³³³³³ᵉ⁻¹ +( -3.33333333333e⁻¹) = 1.0⁻¹² ≠ 0.
    Thank you for trying.