I'll Be Proud If You Solve This

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  • Опубліковано 29 січ 2025

КОМЕНТАРІ • 46

  • @BriTheMathGuy
    @BriTheMathGuy  Рік тому +8

    Check out and consider donating to Connect Me!
    connectmego.org/

  • @cameronbigley7483
    @cameronbigley7483 Рік тому +48

    I find it slightly more intuitive to let a = 5^x and b = 2^x, then you'll get the same factoring situation.

  • @golzaye3393
    @golzaye3393 Рік тому +19

    I got it! This is pretty much the only one I've ever gotten from your channel this quickly.

  • @Unidentifying
    @Unidentifying Рік тому +4

    really nice, using nearly all of the power tricks
    i went with logs in earlier steps and got wrong answer or made some mistake somewhere

  • @zeldalink6278
    @zeldalink6278 Рік тому +44

    This is one of the basic exercises for grade 12 in Vietnam

  • @EdMatthewMorales
    @EdMatthewMorales 3 місяці тому +2

    *ignores complex solution*

  • @lucykitsune4619
    @lucykitsune4619 Рік тому +4

    I tried to divide them up into prime-factors at first, but I did in the end get x = log_5/2 (2)

  • @maccyote7650
    @maccyote7650 Рік тому +11

    damn i hope i could someday solve this types of problems in no time.

  • @HarshSamosa
    @HarshSamosa 10 місяців тому

    I legit didn't expect it to be easy
    You always present us with mind boggling equations!

  • @cheeseburgermonkey7104
    @cheeseburgermonkey7104 Рік тому +1

    I think you could improve the final answer a bit more by using logarithms with whole number bases

  • @RenegadePawn___
    @RenegadePawn___ 10 місяців тому

    Be proud of me, I got this one. Slightly different approach: a = 5^x and b = 2^x. Moving everything to one side gives a^2 + ab -6b^2 = 0, which factors to (a - 2b)(a + 3b) = 0. Since 5^x > 0 and 2^x > 0 we know a + 3b = 0 has no real solution. Therefore a - 2b = 0 gives the solution. 5^x - 2*2^x = 0 leads to the same answer, or since I have weird ln reflexes mine says the equivalent ln 2 / ln (5/2).

  • @opbunte8862
    @opbunte8862 Рік тому +3

    In India this is very basic for a grade 11th students preparing for jee (joint entrance exam )

  • @ANTI_UTTP_FOR_REAL
    @ANTI_UTTP_FOR_REAL Рік тому +2

    Yall saying its challanging meanwhile its the most casual excersise for 11th grade in Armania 😂

  • @thisjosiah
    @thisjosiah 12 днів тому

    I did it with 2^x as the thing to solve for in the quadratic equation with - 25^x as the c term(it cancels out nicely) leaving (5^x +- 5* 5^x)/12 and then it's pretty easy to solve from there

  • @MattHatter-t5r
    @MattHatter-t5r Рік тому +2

    Divide by 4^x solve quadratic x= ln2/ln5-ln2 hope thats right anyway that was a nice question 😉

  • @vishalmishra3046
    @vishalmishra3046 Рік тому

    25^x + 10^x = 6 x 4^x => (5/2)^(2x) + (5/2)^x - 6 = 0 => (5/2)^x = (-1 +/- √(1+24))/2 = (-1 +/- 5)/2 = 2 or -3. Taking logs on both sides, x = ln(2) / (ln(5) - ln(2))

  • @tb0707-m8w
    @tb0707-m8w Рік тому +1

    Awesome video, Brian! What do you think about sharing the manim code of your videos in exchange for a donation on patreon or somewhere else?

  • @SapphireZYT
    @SapphireZYT 8 місяців тому +1

    16^x - 12^x = 9^(x+y)
    3^y = √2
    x=?

    • @Jius-fg5zq
      @Jius-fg5zq 3 місяці тому

      x = (ln(-1 ± sqrt(17)) - ln2)/(2ln2 - ln3)
      Real solution when there is + sign

  • @nguyentien5331
    @nguyentien5331 Рік тому

    I divided both sides by 25^x and let t = (2/5)^x (t>0) and then solved the quadratic equation ( 1 + t = 6t^2 ) then I found t = 1/2 and -1/3 ( which is eliminated because t>0 ), if t = (2/5)^x then (2/5)^x = 1/2 => x = log_2/5 (1/2), that's my way!

  • @tlpiwsool2d
    @tlpiwsool2d 7 місяців тому +1

    are you kidding me? you only have to apply the quadratic formula to find out 5^x=2^(x+1), and then its easy

  • @scottleung9587
    @scottleung9587 Рік тому +1

    I solved it using natural logs (if that's valid).

  • @zafarbaluch8782
    @zafarbaluch8782 3 місяці тому

    Make a video about prime no...an unjusfied formula that is available 😅

  • @meiliodinson
    @meiliodinson Рік тому +8

    log _base 5/2_ ??
    I mean yes but no one uses that. The natural log is a better choice here. x = ln(2)/(ln(5) - ln(2))

    • @Ninja20704
      @Ninja20704 Рік тому +5

      They are the same thing due to the change of base formula. log_5/2 is more direct while ln would be more useful if we actually wanted to calculate an approximation.

    • @meiliodinson
      @meiliodinson Рік тому +1

      @@Ninja20704 I know they're the same thing, my point was that while technically correct, no one uses bases that way. But yeah maybe it's just a habit from the need to often calculate an approximation. Still feel like it's incomplete writing base 5/2 x)

  • @wrestlingworld3410
    @wrestlingworld3410 Рік тому +1

    The answer to this question is 0.4307 to 4 decimal places, as a result of log2/log5=0.4307 that is the correct answer

  • @aryanagrawal9103
    @aryanagrawal9103 Рік тому +1

    Easy

  • @_Spread_Positivity_
    @_Spread_Positivity_ Рік тому

    Here in the first minute of upload

  • @kikivoorburg
    @kikivoorburg Рік тому +1

    I got the answer through a different method (completing the square). This one was fun!
    Here’s my LaTeX in case anybody wants to look, though it’s a bit messy in a comment:
    25^x + 10^x &= 3 \cdot 2^{2x + 1} \\\\
    5^{2x} + 2^x 5^x - 3 \cdot 2^{2x + 1} &= 0 \\
    (5^x)^2 + 2 \cdot 2^{x-1} \cdot 5^x - 24 \cdot (2^{x-1})^2 &= 0 \\
    (5^x + 2^{x-1})^2 &= 5^2 (2^{x-1})^2 \\
    5^x + 2^{x-1} &= 5 \cdot 2^{x-1} \\
    5^x &= 4 \cdot 2^{x-1} \\
    5^x &= 2 \cdot 2^x \\
    x &= \log_5 (2) + x \log_5(2) \\\\
    \therefore x &= \frac{\log_5(2)}{1-\log_5(2)}

  • @RadekBuczkowski-h2y
    @RadekBuczkowski-h2y 11 місяців тому

    This equation has only one real solution, but it has infinitely many
    complex number solutions. Here is how to calculate all solutions, and
    also how to calculate a natural logarithm of a negative real number.
    (25^x + 10^x)/3 = 2^(2x+1)
    (5^x * 5^x) + (2^x * 5^x) = 6(2^x * 2^x)
    5^x * 5^x 2^x * 5^x
    -------------- + ------------- = 6
    2^x * 2^x 2^x * 2^x
    ((5/2)^x)^2 + (5/2)^x - 6 = 0
    t = (5/2)^x
    t^2 + t - 6 = 0
    t = (-1 +- 5)/2
    (t + 3)(t - 2) = 0
    Real solutions:
    (5/2)^x = -3 (no real solutions)
    (5/2)^x = 2
    x = ln(2) / ln(5/2) = ln(2) / (ln(5) - ln(2))
    Complex solutions:
    e^(i*pi) = -1 (Euler's identity)
    e^0 = 1
    C = i*2*pi*n where n E Z
    e^(i*pi + C) = -1
    e^(0 + C) = 1
    (5/2)^x = -3 = 3 * e^(i*pi + C)
    x = ln(3 * e^(i*pi + C)) / ln(5/2)
    x = (ln(3) + i*pi + C) / ln(5/2)
    (5/2)^x = 2 = 2 * e^(C)
    x = ln(2 * e^(C)) / ln(5/2)
    x = (ln(2) + C) / ln(5/2)
    Answer:
    x = (ln(3) + i*pi + i*2*pi*n) / ln(5/2) where n E Z
    x = (ln(2) + i*2*pi*n) / ln(5/2)
    if n = 0:
    x = ln(2) / ln(5/2) (the only real solution)

  • @ryanyeater5669
    @ryanyeater5669 9 місяців тому

    i tried it where did i go wrong
    ((25^x)+(10^x))=3*(2^(2x+1))
    ((25^x)+(10^x))/(2^(2x+1))=3 (2^(2x+1))=((2^2x)*2)
    subsitute to get
    (((25^x)/2)+((10^x)/2)))/(2^(2x)=3
    take 2xth root of it to get
    (((25^x)/2)+((10^x)/2)))to the 2xth root/2= 3to the 2xth root
    *2on both
    ^2x on both
    use (a/2)+(b/2)=((a+b)/2)
    (((25^x)+(10^x))/2)=(2^2x)*3
    /3
    *2
    (((25^x)+(10^x))/3)=(2^(2x+1))
    ((25^x)+(10^x))=3*(2^(2x+1))
    dang it