Oxford University 🎓 Entrance Exam | Can you solve ?

00:03:27
https://www.youtube.com/watch?v=v1B7KHXrKeU

Resumo

TLDRIn this video, the presenter solves the exponential equation x^x = √(x^x) by applying logarithmic properties. The equation is transformed and simplified, leading to two potential solutions. The presenter emphasizes that x cannot be zero and ultimately finds that the solutions to the equation are x = 1 and x = 4. The video encourages viewers to subscribe and engage with the content.

Conclusões

  • 🔍 The equation x^x = √(x^x) is solved using logarithms.
  • 📏 Properties of logarithms simplify the equation.
  • 🧮 The solutions found are x = 1 and x = 4.
  • 🚫 x = 0 is not a valid solution.
  • 📈 Taking the natural logarithm helps isolate terms.

Linha do tempo

  • 00:00:00 - 00:03:27

    The video introduces an exponential equation, x^x = √(x^x), and explains how to manipulate it using logarithms. The equation is transformed into a more manageable form by applying logarithmic properties, leading to a common factor of ln(x). This results in two potential solutions: ln(x) = 0 or √x = x/2. The first solution, x = 1, is derived from ln(x) = 0. The second solution is found by squaring both sides of the equation √x = x/2, leading to x^2 = 4x, which simplifies to x(x - 4) = 0. Since x = 0 is not a valid solution, the final solutions are x = 1 and x = 4. The video concludes with a call to action for viewers to subscribe and engage with the content.

Mapa mental

Vídeo de perguntas e respostas

  • What is the equation being solved?

    The equation is x^x = √(x^x).

  • What method is used to solve the equation?

    The method used involves taking the natural logarithm of both sides and applying properties of logarithms.

  • What are the final solutions to the equation?

    The final solutions are x = 1 and x = 4.

  • Is x = 0 a solution?

    No, x = 0 is not a solution according to the original equation.

  • What happens when you take the natural logarithm of both sides?

    Taking the natural logarithm allows us to simplify the equation and isolate terms.

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Legendas
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Rolagem automática:
  • 00:00:00
    Hello, welcome back once again. Today
  • 00:00:02
    we're going to solve this interesting
  • 00:00:04
    exponential equation. X^x is equal to
  • 00:00:08
    square root of X to the^ of X. Right?
  • 00:00:12
    Let's take note of the following. When
  • 00:00:14
    we have the nth
  • 00:00:16
    root of a raised to the power of B. This
  • 00:00:20
    is same as a^ of b / n. So we're going
  • 00:00:25
    to apply this at the right hand side. So
  • 00:00:27
    our equation will become x^ of roo
  • 00:00:31
    x is equal to x^ of x all over 2. Now
  • 00:00:38
    let us l in both sides but take note of
  • 00:00:40
    this since we know that 0^ of 0 is not
  • 00:00:43
    equal to 1. Then x cannot be equal to z.
  • 00:00:47
    Okay. Now to solve this equation we take
  • 00:00:50
    the ln on both sides. So ln of x^ of x
  • 00:00:54
    is equal to the ln of x^ of x all / 2.
  • 00:00:59
    Now with the property of logarithm the
  • 00:01:01
    log of a^ of b is equal to b * the log
  • 00:01:05
    of a. So this will become roo
  • 00:01:09
    x the ln of x is = x / 2 * the ln of x.
  • 00:01:16
    This term here take it to the left hand
  • 00:01:18
    side. We have square root of x * the ln
  • 00:01:21
    of x
  • 00:01:22
    - x / 2 * the ln of x. This is equal to
  • 00:01:27
    0. Now we can see ln of x is common. We
  • 00:01:30
    take it out. So we have ln of x. Then
  • 00:01:33
    into bracket we're going to have square
  • 00:01:35
    root of x minus x / 2 is equal to 0.
  • 00:01:42
    So from here this implies that either ln
  • 00:01:45
    of x is equal to 0 or roo of x - x
  • 00:01:52
    / 2 is equal to z. So from here take the
  • 00:01:55
    common base of e on both sides. So e and
  • 00:01:58
    ln get cancel here we have x is equal to
  • 00:02:01
    e to the^ of 0 which is 1. So 1 is a
  • 00:02:04
    solution. And for here take this term
  • 00:02:06
    here to the right hand side. it will
  • 00:02:08
    become square root of x = x / 2. Now
  • 00:02:12
    square both sides of this
  • 00:02:16
    equation. So here the square gets cancel
  • 00:02:18
    with the square root and from the right
  • 00:02:20
    hand side we have
  • 00:02:22
    x² / 2 that will give us 4. This is
  • 00:02:26
    equal from the left hand side we have x.
  • 00:02:29
    Now cross multiply x² * x will give us
  • 00:02:33
    x. This is equal to 4 * x will give us
  • 00:02:36
    4x. Oh, I mean this is x square. Sorry
  • 00:02:39
    for that. Now, we know that sol we know
  • 00:02:41
    that zero is not a solution, right?
  • 00:02:44
    Because if you take 4x to the left hand
  • 00:02:46
    side, you're going to factor out x and
  • 00:02:48
    you have something like this. x into
  • 00:02:50
    bracket x - 4 is equal to 0. And here we
  • 00:02:53
    can say x is equal to 0. But according
  • 00:02:55
    to the original equation, 0 is not a
  • 00:02:57
    solution. Therefore, let's kindly divide
  • 00:03:00
    both side
  • 00:03:01
    by. So this will imply that x²id x will
  • 00:03:05
    give us x. This is equal to 4x / x will
  • 00:03:08
    give us 4. So we quickly arrive at the
  • 00:03:12
    solution. So we have two solutions. x =
  • 00:03:16
    1 and x = 4. Thank you for watching.
  • 00:03:21
    Please if you really enjoy the video,
  • 00:03:23
    kindly subscribe. Also like, comment and
  • 00:03:26
    share.
Etiquetas
  • exponential equation
  • logarithm
  • solution
  • mathematics
  • x^x
  • square root
  • properties of logarithms
  • natural logarithm
  • algebra
  • equation solving