I Solved A Nice Exponential Equation

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  • เผยแพร่เมื่อ 4 ธ.ค. 2024
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ความคิดเห็น • 14

  • @justabunga1
    @justabunga1 วันที่ผ่านมา +2

    We could subtract (x-4)^(x+3) to both sides, which we get (x-4)^(x+5)-(x-4)^(x+3)=0. Factor out (x-4)^(x+3), which is (x-4)^(x+3)((x-4)^2-1)=0. Set each factor equal to 0, so (x-4)^(x+3)=0 and (x-4)^2-1=0. (x-4)^(x+3)=0 has one solution, which is x=4. (x-4)^2=1 making this x-4=±1, xo x=3 and x=5. We have all 3 solutions, so therefore, x=3, 4, and 5. If you graph those two solutions and using the calculus method, we only see that there are two intersection points at x=4 and x=5 since x=3 is out of the domain. Therefore, the domain for those two functions has to be x≥4.

  • @rakenzarnsworld2
    @rakenzarnsworld2 วันที่ผ่านมา +1

    x = 5 or 4 or 3

  • @TurquoizeGoldscraper
    @TurquoizeGoldscraper วันที่ผ่านมา +1

    I set y = x - 4 and moved the entire equation to one side to get:
    y^(y + 9) - y^(y + 7) = 0
    y^(y + 7) (y^2 - 1) = 0
    Therefore:
    y^(y+7) = 0 or y^2 = 1
    This gives me y = -1, 0 or +1.
    Going back to x by adding 4 gives:
    x = (3, 4, 5)

  • @scottleung9587
    @scottleung9587 วันที่ผ่านมา +2

    I got x=3,4,5 as solutions.

  • @Quest3669
    @Quest3669 วันที่ผ่านมา +1

    (X-4)^x+3[ (x-4)^2-1]= 0 gives x= 4; 5; 3 solns.

  • @etianslab5239
    @etianslab5239 วันที่ผ่านมา

    1^n=1 so if x-4=1, then x=5, easy peasy lemon squeezy

  • @neuralwarp
    @neuralwarp วันที่ผ่านมา

    You messed up the x=4 case

    • @SyberMath
      @SyberMath  วันที่ผ่านมา

      Why?

    • @SrisailamNavuluri
      @SrisailamNavuluri วันที่ผ่านมา

      ​@@SyberMaththe base was taken -1,0,1 where it is other than 0.

    • @alexandervolok1863
      @alexandervolok1863 วันที่ผ่านมา

      @@SyberMath 0^(4+3) = 0^(4+5) but you wrote 0^3 = 0^5. Doesn't matter btw, but a bit inaccurate

  • @Don-Ensley
    @Don-Ensley วันที่ผ่านมา +2

    problem
    (x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺⁵⁾
    (x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺ ³ ⁺ ²⁾
    (x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺ ³⁾(x-4)²
    Subtract.
    (x-4)⁽ˣ⁺³⁾ - (x-4)⁽ˣ⁺ ³⁾(x-4)² = 0
    Factor.
    (x-4)⁽ˣ⁺ ³⁾ [ 1 - (x-4)² ] = 0
    By zero factor property,
    (x-4)⁽ˣ⁺ ³⁾ = 0
    x = 4
    1 = (x-4)²
    = x²- 8x + 16
    x²- 8x + 15 = 0
    x = 5, 3
    answer
    x ∈ { 3, 4, 5 }

    • @musicsubicandcebu1774
      @musicsubicandcebu1774 วันที่ผ่านมา

      Thanks for taking the time to use superscripts.

    • @tunistick8044
      @tunistick8044 วันที่ผ่านมา +1

      you could've really just did:
      (x-4)²=1 x-4=±1 x=±1+4 x=3 or x=5

    • @SyberMath
      @SyberMath  วันที่ผ่านมา +1

      Nice!