How to solve exponential equations

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  • Опубліковано 30 січ 2025

КОМЕНТАРІ • 74

  • @albertoesquivias6073
    @albertoesquivias6073 10 місяців тому +49

    Please do more of these "how to solve" type of videos and I hope it becomes a series, it was really easy to follow

    • @bprpmathbasics
      @bprpmathbasics  10 місяців тому +19

      I plan to. Glad to hear they are helpful! Thanks.

    • @kingsan3162
      @kingsan3162 8 місяців тому

      @@bprpmathbasics Hello,thank you so much and i consider you a teacher of mine even though i never met you live.
      But i wanna ask you if you can do 2 examples of 1 equation.
      Becuase in that case it helps anyone see the bigger picture.

  • @TerraBlo
    @TerraBlo 10 місяців тому +42

    12:20
    I love how he shows these random bloopers

    • @lotaniq4449
      @lotaniq4449 10 місяців тому +1

      its an accident lol

    • @TerraBlo
      @TerraBlo 10 місяців тому

      @@lotaniq4449 he does it quite a lot

    • @TerraBlo
      @TerraBlo 10 місяців тому

      @@lotaniq4449 he did mean it he puts it on some other videos sometimes

    • @tfg601
      @tfg601 5 місяців тому

      I don't get it

  • @euloge996
    @euloge996 10 місяців тому +28

    12:19 lazy editing😂
    love you bro

  • @pedrogarcia8706
    @pedrogarcia8706 10 місяців тому +11

    11:15 you can also use ln(pos)/ln(b). most calculators with have natural log in addition to log base 10.

    • @ronaldjensen2948
      @ronaldjensen2948 10 місяців тому

      For me this was more intuitive. we just go from x ln(5) = ln(2) to x = ln(2)/ln(5)

    • @Erlewyn
      @Erlewyn 10 місяців тому +1

      That's actually how I was taught 20+ years ago, in high school in France. I don't think you even learn about different bases log before college here.

  • @TheZmoliver
    @TheZmoliver 8 місяців тому +2

    You take the stress out of math, man. Thank you!

  • @the_m_original
    @the_m_original 3 місяці тому +2

    for question 9, here is the imag solution:
    e^x = -3
    x = ln(-3) ; ln both sides
    x = ln(3*-1) ; factorize -3
    x = ln(3) + ln(-1) ; split
    x = ln(3) + ln(e^iTheta) ; let e^iTheta = -1, and Theta be equal to Pi + 2nPi where n is any positive integer
    x = ln(3) + i(Pi+2nPi)

  • @bencepaul3497
    @bencepaul3497 3 місяці тому +1

    Thanks so much for your videos! I really get a lot out of them!!

  • @DrTWG
    @DrTWG 9 місяців тому +1

    You are teaching a lot of people some great maths skills .

  • @epicstar86
    @epicstar86 10 місяців тому +7

    for question 6 I used exponential properties before logarithmic ones

  • @fireball54110
    @fireball54110 10 місяців тому +105

    -okay, number 6-

    • @nicholaslama6670
      @nicholaslama6670 10 місяців тому +20

      Bro, I got so confused, I thought I was having some recurring déjà vu's

    • @idontcareforthis
      @idontcareforthis 10 місяців тому

      @@nicholaslama6670No déjà-vu for me, but I was on the verge of cursing at my phone for glitching.

    • @charg1nmalaz0r51
      @charg1nmalaz0r51 9 місяців тому +8

      @@nicholaslama6670 got stuck in the worst version of groundhog day 🤣

  • @sankaranbs4024
    @sankaranbs4024 10 місяців тому +40

    I thought, I have I Internet problem with 'Okay, number 6 what's e though'

  • @leonardobarrera2816
    @leonardobarrera2816 10 місяців тому +2

    Exponential equations are so easy that I have never saw on my college and I can solve all of them!!!

  • @HimanshuYadav-v3r
    @HimanshuYadav-v3r 9 місяців тому +2

    i am promoted to class 10th 2 months ago and i solved all. I cant believe it.

  • @gigi4874-w3w
    @gigi4874-w3w 10 місяців тому +1

    Fantastic review! 🙏🏽👏🏽

  • @RedoanaSultana
    @RedoanaSultana 10 місяців тому +1

    Question 7:
    3^(x-2) = 5^(x+4)
    (3/5)^x = 5^4 * 3^2
    x = log(5^4 * 3^2)/log(3/5)
    Question 8:
    7^(2x-1) = 2^(4x+3)
    (7/4)^(2x) = 2^3 * 7
    2x = log(2^3 * 7)/log(7/4)

  • @Fenamer
    @Fenamer 10 місяців тому +3

    Can you make more ''How to solve' videos? I think these videos are easy to follow through, and I want more so I can learn bits of precalc and calc.

  • @suchaluch5615
    @suchaluch5615 10 місяців тому +2

    Great presentation!
    Thanks

  • @fromscratch8774
    @fromscratch8774 7 місяців тому +1

    For 9: e^2x + e^x - 6 = 0
    why wouldn't collecting like terms and using ln on both sides work?

  • @Gremriel
    @Gremriel 10 місяців тому +2

    Most of what you talk about goes over my head, but occasionally I learn something.

  • @klmcwhirter
    @klmcwhirter 10 місяців тому +2

    Love the video.
    But I saw you miss this a couple of times now.
    The reason the exponential property of same bases works is because if you take the log of both sides they cancel. _Note I saw you covered this later, but it would be helpful to explain in problem 1._
    2^(3X+1) = 2^5
    log₂ 2^(3X+1) = log₂ 2^5 (how do I do strikeout in a comment?)
    Which leaves you with ...
    3X+1 = 5
    Heads up - that extra explanation will make things clearer for some students.
    Great job though. I love your videos!

    • @cheeseplated
      @cheeseplated 10 місяців тому +2

      -for strikethrough- put - on either end of the text
      _for italics_ it's _
      *for bold* it's *
      *_-and you can combine them-_*

    • @bprpmathbasics
      @bprpmathbasics  10 місяців тому

      Thank you!

    • @Apollorion
      @Apollorion 10 місяців тому

      The reason that that 'exponential property of same bases' works is that:
      A: the base is bigger than 0 as well as well unequal to 1
      and
      B: if y=a^x and z=dy/dx then z is unequal to 0 and has a constant sign (i.e. d( z/|z| )/dx = 0 no matter the value of x) which implies that for an increase in x, y is always bigger (for a>1) or smaller (for a

  • @DominickRiesland
    @DominickRiesland 10 місяців тому +3

    What do you do when the teacher tells you to express the answer to #7 and #8 as a single logarithmic function? 😂

    • @bprpmathbasics
      @bprpmathbasics  10 місяців тому +1

      Show them this video: ua-cam.com/video/WVGZbO0esOM/v-deo.html
      : )

  • @axn_anime
    @axn_anime 7 місяців тому +1

    Can you please do one for trigonometry 🥺❤️

  • @ghostkiller4071
    @ghostkiller4071 10 місяців тому +3

    Thanks for everything

    • @penguin9257
      @penguin9257 10 місяців тому

      This sounds concerning

  • @GirishManjunathMusic
    @GirishManjunathMusic 10 місяців тому +3

    Before watching the video:
    Question 1:
    2↑(3x + 1) = 32
    We know that 32 = 2⁵
    2↑(3x + 1) = 2⁵
    Comparing like terms:
    3x + 1 = 5
    3x = 4
    x = 0.75.
    Question 2:
    27↑x = 9↑(2x - 3)
    27 = 3³ and 9 = 3²:
    3↑3x = 3↑(4x - 6)
    Comparing like:
    3x = 4x - 6
    x = 6.
    Question 3:
    5↑(x² + 3x - 4) = 1
    But 5⁰ = 1:
    5↑(x² + 3x - 4) = 5⁰
    Comparing like:
    x² + 3x - 4 = 0
    x² + 4x - x - 4 = 0
    (x + 4)(x - 1) = 0
    x = -4, x = 1.
    Question 4:
    (√2)↑(x + 4) = ⅛
    But ⅛ = 8↑(-1), 8 = 2² and √2 = 2↑½:
    2↑(½x + 2) = 2↑(-3)
    Comparing like:
    ½x + 2 = - 3
    ½x = -5
    x = -10.
    Question 5:
    5↑x = 2
    xln5 = ln2
    x = ln2/ln5.
    Question 6:
    20e↑(3x - 2) = 1200
    e↑(3x - 2) = 60
    3x - 2 = ln(60)
    x = ⅓(ln(60) + 2).
    Question 7:
    3↑(x - 2) = 5↑(x + 4)
    (x - 2)ln3 = (x + 4)ln5
    xln3 - 2ln3 = xln5 + 4ln5
    x(ln3 - ln5) = 2ln3 + 4ln5
    x = (2ln(3) + 4ln(5))/(ln(3) - ln(5)).
    Question 8:
    7↑(2x - 1) = 2↑(4x + 3)
    (2x - 1)ln7 = (4x + 3)ln2
    2xln7 - ln7 = 4xln2 + 3ln2
    x(2ln7 - 4ln2) = (3ln2 + ln7)
    x = (3ln2 + ln7)/(2ln7 - 4ln2).
    Question 9:
    e↑2x + e↑x - 6 = 0:
    Setting e↑x = t → e↑2x = t²:
    t² + t - 6 = 0
    t² + 3t - 2t - 6 = 0
    (t + 3)(t - 2) = 0
    t = -3 or t = 2
    But as t = e↑x; t > 0:
    t = 2 → e↑x = 2
    x = ln2.
    Question 10:
    2↑x + 3·2↑(-x) = 4
    Setting 2↑x = t → 2↑(-x) = 1/t:
    t + 3/t = 4
    t - 4t + 3 = 0
    t - 3t - t + 3 = 0
    (t - 1)(t - 3) = 0
    t = 1 or t = 3.
    But t = 2↑x:
    Case 1: t = 1:
    2↑x = 1
    x = 0.
    Case 2: t = 3:
    2↑x = 3
    xln2 = ln3
    x = ln(3)/ln(2).

    • @Apollorion
      @Apollorion 10 місяців тому

      I think you made a pair of typo's in Question 10; after multiplying the whole equation with t, t does not turn into t² whereas it should and is also processed as.

    • @GirishManjunathMusic
      @GirishManjunathMusic 10 місяців тому

      @@Apollorion so I have! The radical must not have entered properly, I'm typing these on the phone keyboard, so those radicals are sometimes difficult to enter!

    • @Apollorion
      @Apollorion 10 місяців тому

      @@GirishManjunathMusicWell, bless UA-cam. /s
      If I made a 'mistake' from the perspective of the 'algorithm' then I want to know how/why and not a brutal disappearance without explanation; I want to learn from my mistakes to avoid to repeat them.
      P.S.
      b.t.w. you might help yourself with copy & paste or describing t squared with other symbols e.g. t^2 or tt.

    • @thewendigo9263
      @thewendigo9263 9 місяців тому +1

      You made a mistake on question one. X is 1.33

    • @Apollorion
      @Apollorion 9 місяців тому

      @@thewendigo9263I agree with the mistake you seem to notice (i.e. 3x=4 => x=4/3) but not with the alternate solution you propose i.e. 1.33
      _It is more accurate but still not correct_ . The correct solution in expressed in decimals would finish with an infinite amount of 3s after the dot, not just two.

  • @javier-kk4eq
    @javier-kk4eq 3 місяці тому +1

    here 7 months later great video

  • @Phi1618033
    @Phi1618033 Місяць тому

    Legend says BPRP is still starting example 6.

  • @cdkw8254
    @cdkw8254 10 місяців тому +1

    Where is the 25 find the range of function questions? You said you would upload that in your 25 trig questions video.

  • @lucaspheng6009
    @lucaspheng6009 3 місяці тому +1

    Why not just use the division rule for the logs with negatives?

  • @fullc0de
    @fullc0de 10 місяців тому +2

    @4:46 OMG, I know that 5^0 = 1. How did I not see that I needed to use that when I tried the problem before I watched you solve it?????

    • @darranrowe174
      @darranrowe174 10 місяців тому +1

      Sometimes these facts are so simple, we miss them.
      When I was younger, I know I often saw equality as a one way thing so I could apply it one way but never thought about applying it the other. This is why videos like this can be good. It makes you think of things in ways you never thought of before.

    • @fullc0de
      @fullc0de 10 місяців тому

      @@darranrowe174 When you do make the connection and see how the puzzle piece fits, it's such a great feeling, no matter how simple the insight is! When you don't, it's a total Homer Simpson '"D'oh!" moment! The drama of math.

  • @barryspindler8342
    @barryspindler8342 8 місяців тому +1

    excellent job and done at the right pace. I like your work.

  • @rangga2397
    @rangga2397 9 місяців тому +1

    5:16 i factor (x-3) (x-1) am i wrong?

    • @zilovesmaths
      @zilovesmaths 9 місяців тому +3

      Ah you've done it the other way around
      To factor it you need 2 numbers that multiply to give c (-4)
      And they need to add to give b (+3)
      You've got 2 numbers that multiply for b and add for c haha
      So if you expand (x-3)(x-1)
      That gives x²-4x+3 instead of x²+3x-4

  • @GigachadHGDYDBHDB
    @GigachadHGDYDBHDB Місяць тому

    x=ln 2
    x=ln 3+iPi

  • @SuperTommox
    @SuperTommox 10 місяців тому +3

    Can you solve log(x)=x?

    • @cdkw8254
      @cdkw8254 10 місяців тому +1

      e^x

    • @cpu_1292
      @cpu_1292 10 місяців тому

      You raise x by both sides to get x^log(x) = x^x
      The log and the x cancels out so you get x = x^x
      x has to be one because 1 = 1^1
      Edit: I made a mistake, there is no real solution
      Edit2: here is the real solution
      Raise 10 by both sides and get 10^log(x) = 10^x
      The 10 and the log cancel out to get x = 10^x so there is no real solution

    • @stokmlnes-flame2025
      @stokmlnes-flame2025 10 місяців тому +1

      @@cpu_1292 i think thats wrong bc log(1)=0 [10^0=1]

    • @stokmlnes-flame2025
      @stokmlnes-flame2025 10 місяців тому +1

      no real solutions

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 10 місяців тому +1

      Not in the real plane. This can be shown by knowing ln(x)=x is the same as x=e^x.
      x2>1 means e^x>2^x, so e^x>x, so x>ln(x).

  • @aryanbehera1772
    @aryanbehera1772 Місяць тому +1

    As a 7th grader,i solved it till 4th😅😅 is it fine?

  • @kloneo
    @kloneo 9 місяців тому +2

    It would be nice if people could download a questions sheet of the shown equations but with the numbers changed. To think at home and rewatch your explanation to complete the question. Make a +2 a +5. Something like that for the video questions. The next video give the answers as a 2 sec show frame before the new vid. Mayby win something, Or on patreon or something.

  • @DokkanPapaya
    @DokkanPapaya 2 місяці тому +2

    So this and that cancel

  • @fizisistguy
    @fizisistguy 4 місяці тому +2

    This was clearly not a basic to hard one.
    😅

  • @euloge996
    @euloge996 10 місяців тому +2

    i love you brro

  • @naming_is_harddd
    @naming_is_harddd 10 місяців тому +3

    Can you please upload more on your main channel? I miss the exciting content on that channel

  • @ssdrxgod
    @ssdrxgod 10 місяців тому +1

    ~I Found All Of Them Ez

  • @donwald3436
    @donwald3436 10 місяців тому +1

    and as always that's it lol.