Is this equation solvable? x^ln(4)+x^ln(10)=x^ln(25)

[@blackpenredpen]

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[@jonathan3488]

Your ability to change colors quickly doesn't cease to amaze me.

[@ryboi1337]

Fr. So seamless

[@gubunki]

also erasing half the board with a tap is a nice skill

[@nut4ku]

Average asian mathematics lecture's muscle memory

[@savitatawade2403]

😂😂​@@gubunki

[@savitatawade2403]

​@@nut4ku wait hes asain??

[@pranavrs184]

I tried it, but came up with a different method: x^ln4 + x^ln10 = x^ln25 a = x^ln5 , b = x^ln2 b^2 + ab = a^2 Using Quadratic Formula: a = b(1 + root5)/2 [since a and b must be the same sign, the other one is ignored] x^ln5 = x^ln2(phi) [dividing by x^ln2, means x^ln2 must not be 0. So 0 is a Solution. 😅 Didn't realize at first.] x^ln(5/2) = phi x = phi^(1/ln(5/2)) Edit: Essentially I found out after watching the video for his solution that phi^(1/(ln(5/2)) = phi^(log(base5/2)(e)) = e^log(base(5/2))(phi). So I got the Answer!!! Woohoo!!

[@thangnguyen-iw8tb]

You should divide both side by b^2. And then use quadratic formulas with (A/b)^2+(a/b)-1=0

[@pranavrs184]

Both end up with the same relation a and b. b comes out common anyway for the relation. If I had divided by b^2, I would have gotten a/b = phi. It is just my instinct to use it without dividing, as I had practiced all along.

[@scottleung9587]

That was my method too!

[@youngsandwich9967]

I did it this way too (except I made b=x^ln5 and a=x^ln2). Got phi^(1/(ln5-ln2)) which is the same thing.

[@thefireyphoenix]

sameee

[@atnernt5196]

Been a long time since I have seen a math equation and problem that was just genuinely cool. Having phi show up was a real nice surprise.

[@keylime6]

The golden ratio caught me off guard, but the fact that it’s there is amazing

[@jamescollier3]

2:50 Voodoo

[@ivan1793]

This was my idea: x^ln4 + x^ln10 = x^ln25 Leave out the case x=0 and divide both sides by x^ln10. Distributing the denominator and applying properties of the power you get x^(ln4-ln10) + 1 = x^(ln25 - ln10) Applying properties of logarithms we get x^ln(2/5) + 1 = x^ln(5/2) Substituting z=x^ln(5/2), and noting that ln(2/5)=-ln(5/2) the equation becomes z^-1 + 1 = z Which can easily become a quadratic equation and the rest follows in a similar way to what you did.

[@dnd2008yi]

Got it in 2 min buddy!! You definitely helped me a lot to prepare logarithms for competitions..! Gratitude from India 🇮🇳

[@Gezraf]

what i did was move everything to one side and factor x^ln4(x^ln25/4 - x^ln10/4 -- 1) = 0 --> x^ln25/4 is just x^ln10/4 squared so in this case u can let t = x^ln10/4 leading to the quadratic equation t^2 - t - 1 = 0 --> t solves for (1+-sqrt(5) / 2 --> x^ln10/4 = (1+--sqrt(5) / 2 --> take ln() on both sides --> ln10/4*lnx = ln((1+--sqrt(5)/2)) --> lnx = ln((1+--sqrt(5)/2)) / ln10 --> to solve for x, e^ on both sides to cancel the lnx to get x --> x = e^(ln((1+--sqrt(5)/2)) / ln10) --> x ≈ 1.69

[@henridelagardere264]

Seems like we're finally back at the golden ratio we all love so much - one BPRP video a week.

[@memebaltan]

He’s posting daily on bprp math basics

[@henridelagardere264]

@@memebaltan Herzlichen Dank!

[@SeekingTheLoveThatGodMeans7648]

Wolframalpha couldn't figure out the given exact form solution from the original form, but could when the equation was transformed.

[@danieluman4793]

I first converted all the ln() functions from: ln(4), ln(10), ln(25) to 2ln(2), ln(2) + ln(5), 2ln(5) Then I divided everybody by x^2ln(5), getting: x^(2ln(2) - 2ln(5)) + x^(ln(2) + ln(5) - 2ln(5)) = 1 Which becomes: u^2 + u - 1 = 0, where u = x^(ln(2) - ln(5)) = x^(ln(2/5)) though my final solution does not look as elegant: x = ((-1 + √5) / 2) ^ (1 / ln(2/5)) Pretty cool to see how many ways there are of solving this equation!

[@TheFrewah]

Yes, cool solution

[@DDroog-eq7tw]

You can also start by dividing everything by x^ln(4) and immediately getting a quadratic of x^ln(5/2). Nice equation.

[@GhostHawk272]

True!

[@TheLukeLsd]

Fiz desta forma.

[@rithvikarun7112]

Exactly what I did

[@martingibbs8972]

Yes. I took it out as a factor.

[@jimschneider799]

Great solution development. I never would have thought to approach it that way, because I never thought about the fact a^ln(b) = b^ln(a). I mean, it's obviously true, but not a tool I would have thought to use.

[@kthwkr]

FINALLY!! A youtube math exercise that I couldn't do in my head. Nice. It brought back some memories of forgotten log properties. Oh, to be a freshman in college again.

[@enric314]

Very good video!👏👏 Anhother way to isolate the X in the last part would be using the NOTE again, and we get X=(phi)^(1/ln(5/2))

[@Chinese_cunt]

x=φ^(1/ln(²/⁵)) ? That's a nice solution, ngl

[@timothybohdan7415]

I got the same thing. X=(phi)^(1/ln(2.5)). Most people don't have a "log to the base 2.5" button on their calculator, so getting rid of log to base 2.5 simplifies to something one can plug into a calculator, where phi is the golden ratio constant, which is equal to [(1+sqrt(5)]/2.

[@mostafakhaled9702]

Try to solve this question that I found in a calculus textbook (by James Stewart): Use a quadruple integral to find the hypervolume enclosed by the hypersphere x^2+y^2+z^2+w^2=r^2 in R4 (I wish I made that up lol)

[@blackpenredpen]

😮

[@BeattapeFactory]

very nice equation and explanation

[@mathemateric]

I noticed immediately that it would form a quadratic equation when we divide through either of X^ln4 and X^ln25 But I didn't know about the switch of X and the argument of the log functions. I really love your videos thanks so much.

[@59de44955ebd]

As a side note, basically the same approach also works without turning it into an exponential equation, after excluding the solution x=0 just divide the original equation by x^(log 4) on both sides, which gives: 1 + x^log(5/2) = x^log(25/4) = x^(log(5/2) + log(5/2)) = (x^log(5/2))^2 Using substitution u = x^log(5/2) gives 1 + u = u^2, and therefor u = phi. By solving for x we then get the same result

[@cdkw8254]

I love how you are able to bring golden ratio everywhere!

[@mapron1]

4 10 and 25 chosen deliberately for that; it's not an accident

[@alro3553]

You can write x^ln4=x^ln(2*2)=x^(ln2+ln2)=x^ln2*x^ln2. Doing soemthing similar for the other terms one finds: x^ln2*x^ln2+x^ln2*x^ln5=x^ln5*x^ln5 Dividing everything by ln2*ln5 x^ln2/x^ln5 + 1 = x^ln5/x^ln2 Define a=x^ln2/x^ln5 Then a+1=1/a Which yields a=phi Then x=phi^(ln5/ln2)=phi^ln(5/2) Which is what you got expressed differently

[@pranavrs184]

I also did the same method, and I posted my take but I actually got phi^(1/ln(5/2)). If a + 1 = 1/a, U get a = ((-1 +- root(5))/2). Which is not phi. U can't take negative. so it will be 1/phi. and answers will match.

[@alro3553]

@@pranavrs184 aahhhh you are complitely right my bad!!

[@pranavrs184]

:)

[@aesthetics_ai]

used quadratic too but i tried a different method, i write ln4 as ln2 +ln 2 , ln 10 as ln2+ ln2 +ln2.5 and ln 25 as ln2 +ln2+ ln2.5+ ln2.5 so i had x^ ln 2 x ^ ln2 + x ^ ln 2 x^ ln 2 x^ln2.5 = x^ ln 2 x^ln 2 x^ ln2.5 x^ln2.5 , so i eliminated x^ln2 x^ln 2 for each term with substracing and was left with 1+ x^ ln 2.5 = (x ^ ln2.5 )^2 , so x^ln2.5 rewrite as t and t^2 - t -1 =0 become my new equation so i solved it as quadratic and in in the end it was x^ln2.5 = 1+- sqrt 5 / 2 and solved for X by ln-ing both sides and it became ln x = 1+- sqrt 5 / 2 / ln 2.5 , going with e final soultion was x= e^ ((ln [1+-sqrt 5/ 2] )/ ln 2.5)) .

[@paulgillespie542]

Brilliant is a great choice for sponsor. It ethically agrees with what you are really trying to do .

from Morocco all my respects...thank you for those genious ideas..i shared this video on facebook

[@atrus3823]

I gave this baby a try, and was surprised I could get the answer (got an imaginary one as well), but I did this a little differently. I noticed the common factors 4 = 2*2, 10 = 2*5, and 25 = 5*5, so I knew there was some way I could leverage that, so I factored it, and with some playing around I got: x^2 ln 2 + x^ln 5* x^ln 2 - x^2 ln 5 = 0. At this point, I noticed that we had (x^ln 2)^2 in the first term, and x^ln 2 in the second, so used the quadratic formula (solving was interesting but too long for KZbin comment) to solve for x^ln 2, which gives x^ln 2 = x^ln 5 (-1 +/- sqrt(5))/2. Then, I moved the x^ln 5 to the other side and combine to get x^ln(2/5) = (-1 +/- sqrt(5))/2. Now I just take the weird root of both sides and I get the same answer.

[@n16161]

You’re lying you didn’t do that. Don’t make up stuff on the internet. Your story doesn’t make any sense. It’s actually insane. No sane person could believe it. There is no way you figured it out and there’s nothing you could say that would make the internet believe you. “Oh, but I explained it,” you’ll say. And I’ll say, “No, I don’t accept your explanation and I think you’re lying.” Who’s with me??! Let’s let this guy hear it!! We don’t believe a word he’s saying!

[@atrus3823]

@@n16161 that's up to you, bubs. Really makes no difference to me.

[@Gusttz20i]

That property it's incredible!

[@HafeezShaikh-w3y]

Answer is nearly equal to 1 + ln(2)

[@Zonnymaka]

You're a phisic, aren't you?

[@michaelbaum6796]

Very nice equation. Thanks for your perfect presentation. Great, as always👌your videos are so great!

[@adityagidde1688]

when i first saw the question in the thumbnail I thought of relating it to the technique of solving integrals by using log there I used the same way x^ln(4)+x^ln(10)=x^ln(25) taking log on both sides lnx^ln(4)+lnx^ln(10)=lnx^ln(25) by ln property lnx^a = a ln x ln(4)*ln x+ln(10)*ln x=ln(25)*ln x canceling out ln x from both sides....we get ln(4)+ln(10)=ln(25) and after reaching here I was like what do we have to find in this in the first place!!🥲

[@johnporter7915]

How did you know to do the step at 5:47 where you set one side equal to the quadratic formula value

[@nahblue]

I love every video that starts with Let's do some math for fun. Thanks!

[@thesnackbandit]

Love it, thanks.

[@shmuelzehavi4940]

This equation may be solved without using the transformation: x^ln(a) = a^ln(x) however, it makes it more convenient. Very interesting clip and nice explanation.

[@stabbysmurf]

That is a really cool problem and solution.

[@keithmasumoto9698]

Difference of two squares and then divide each factor by x^(ln2) giving 2=x^(ln5)

[@mpperfidy]

Hey! Have there always been a warehouse full of colored markers in your teaching studio? I've watched all your videos and have never noticed that feature. Either way, nice touch, like all you do.

[@doctorb9264]

Very cool problem.

[@michelesiosti7461]

With a different procedure I also found two complex conjugate solutions approximately x=-0.567 ± 0.167·i

[@9adam4]

You can avoid using fractional bases by reversing the exponent a second time near the end: x^(ln(5/2)) = phi So x = phi^(1/ln(5/2)) = 1.691...

[@kingyodah5415]

You switch colors just as smoothly as you switch log bases 😄. And that stock of pens in corner..😄

[@DanoshTech]

Can we just acknowledge the shear number of markers in the bottom right bro be spending 25 hours a day on math to use them up

[@akivaschwartz3255]

Sheer

[@erichury]

He needs an expo sponsorship

[@DanoshTech]

@@erichury indeed

[@thirstyCactus]

Awesome! For the solution, why is e^log(base 5/2)(phi) preferable to phi^(1/(log(5/2)), or phi^log(base 5/2)(e)? ❤

[@azizbronostiq2580]

I dont understand half of the video but it's still fun to watch

[@victorchrist9899]

Nicely done. ❤

[@romain.guillaume]

For the first time in a while I impressed myself. I looked at the exponent, told myself, 4=2*2, 25=5*5 and 10=2*5 that would be nice to divide either by 4 to have a 5/2 and (5/2)**2 appear. Then I looked at the signs in front of the coefficient and I said to myself : I don’t know how but there is the golden ratio hidden in this equation.

[@thejaegerbomber99]

I solved it using only power properties, but I got phi^ln(2/5), but then realized that, doing some additional operations, I got the same result as in the video.

[@redpepper74]

I think it should be phi^(1/ln(5/2))

[@lumina_]

wow that's cool

[@professorrogeriocesar]

Very good. Cheguei nessa resposta equivalente: [ (sqrt(5-1)/2 ] ^ (1 / ln(2/5) ).

[@johnathaniel11]

This guy is always sponsored by brilliant. He probably has a free course by now from them 😂 I love learning from him. He’s the whole reason I was so interested in integrals

[@richardfredlund8846]

in chess there is something called coursera where many of the top players make courses. (e.g. on openings or whatever) . @fourthofno9184 I don't know how brilliant works but your comment made me think, if blackpenredpen could do a course for them.

[@andrewstrom8157]

So I looked at the nonzero solution to x^(ln(9)) + x^(ln(12)) = x^(ln(16)) and noticed the solution was e^(log base 4/3 (phi)). So this leads to the idea that maybe for positive values a and b the nonzero solution to x^(ln(a^2)) + x^(ln(ab)) = x^(ln(b^2)) is e^(logbase b/a (phi)). I'm going to work this out to see if it is true.

[@golgondaDesert]

I tried to do it in another way by manipulating around the powers but only got to 0, how do I know if an equation like this one has more than 1 solution ? and is there a methodology I could follow to find these solutions? or do I just have to study extremely hard maths to become able to find them

[@sultanwiranatakusumah4154]

Thanks

[@theupson]

the first transformation isn't material. you can just divide by x^log(25) and simplify: x^(2log(2/5))+x(log(2/5))=1 and you're on the same line.

[@romanbykov5922]

wonderful!

[@lukelu8042]

brilliant!

[@Anandbhaai]

Love your videos

[@LinnDLuffy]

Can u help solve Integral of 1/(Square root of x(to the power 3)+x)

[@DARKi701]

The "equation of the year" definition reminds me when I participated to the local math contest which I participated when I was 12

[@kerenelbaz2607]

nice and very easy

[@Nxck2440]

Starting with: x^(ln 4) + x^(ln 10) = x^(ln 25) Divide both sides by the RHS: x^(ln 4) / x^(ln 25) + x^(ln 10) / x^(ln 25) = 1 Use law of indices: x^(ln 4 - ln 25) + x^(ln 10 - ln 25) = 1 Use law of logs: x^(ln 4/25) + x^(ln 10/25) = 1 Form a quadratic: (x^(ln 2/5))^2 + x^(ln 2/5) - 1 = 0 Solve quadratic: x^(ln 2/5) = (-1 +/- sqrt(5)) / 2 Since powers are always positive, choose + solution only: x^(ln 2/5) = (-1 + sqrt 5) / 2 Therefore x = ((-1 + sqrt 5) / 2)^(1 / ln 2/5) = 1.69075... Getting my answer to match the form in the video was the hardest part! let phi = (1 + sqrt 5) / 2, then 1/phi = (-1 + sqrt 5) / 2 Then x = (1/phi)^(1 / ln 2/5) x = phi^(-1 / ln 2/5) x = phi^(1 / ln 5/2) x = phi^(ln e / ln 5/2) x = e^(ln e / ln 5/2 * ln phi) x = e^(ln phi / ln 5/2 * ln e) x = e^(log_{5/2}(phi) * ln e) x = e^(log_{5/2)(phi))

[@JakubS]

What's cool is that e≈5/2, so the log with base 5/2 is approximately ln, so the solution is approximately the golden ratio.

[@sanay-gt9pl]

I used the same property but then used graphs to find the no of soln

[@vishumathematics123]

Thankyou sir,this is goos way.

[@MwiibaKuyokwa]

you are the best

[@Fred-yq3fs]

Always the same trick with those: divide the 2 sides by smth clever so you end up with a ratio and its inverse. The 1st step is to decompose the exponents with the ln rules ln4=ln2+ln2, etc... Divide both sides of the equation by x^ln2*x^ln5, and you get to 1/X+1=X with X =x^ln2.5 X = phi (well known golden ratio), the negative root is rejected given X>0 x = Exp(lnPhi / ln(2.5)) year 11 stuff.

[@sfbefbefwfvwfvsf2722]

this is brilliant.

[@ZapOKill]

I like the stash of markers

[@mtaur4113]

x=e^a, 4^a + 10^a = 25^a s=2^a, t=5^a s^2+st-t^2=0 Hmmm, can solve s as a function of t or vice versa as a quadratic, maybe the equation is solvable for a when you substitute back. Perhaps a log or W function will show up later if the problem is nice enough to allow it.

[@mtaur4113]

And then I start watching and basically 5/2 and its square are nicely there already, to the a power. I wonder how it goes if I follow through on what I was going to do. Equivalently, divide by t^2, and s/t=w. Or do it the other was around with s^2.

[@dannydewario1550]

​@@mtaur4113 Try to factor s^2 + st - t^2 by imagining it in its factored form with injected variables like this: (s + ut)(s - vt) = 0 We can expand this to get something similar to what we started with: s^2 + (u - v)*st - uv*t^2 = 0 If we want this new formula to end up just like our original formula, then both 'u' and 'v' must have values which satisfy these two equations: 1) uv = 1 (comes from uv*t^2) 2) u - v = 1 (comes from (u - v)*st) Solving these two simultaneous equations yields us with: u = (1 + sqrt(5)) / 2 v = (-1 + sqrt(5)) / 2 We can rewrite this with the golden ratio 'phi': u = phi v = 1 / phi Now we can substitute 'u' and 'v' back into our factored formula: (s + (phi)*t)(s - (1 / phi)*t) = 0 My thumbs are getting tired typing all this on my phone lol, so I'll stop here. But just substitute back in 2^a and 5^a for 's' and 't', and you should be able to solve for a.

[@mtaur4113]

@@dannydewario1550 Kind of nice, probably was better just to observe the 5/2 and (5/2)^2 in the first place, but why quit halfway if it's doable?

[@dannydewario1550]

@@mtaur4113 Exactly! Plus it was just nice to see someone else in the comments who approached it similar to how I did. I also didn't think about using the quadratic formula with 5/2. This definitely took more steps than the solution in the video, but it's cool that there's more than one method of solving.

[@simongorka7132]

I got ((1+sqr5)/2)^(1/ln(5/2)), so basically the same thing :D

[@AcTpaxaHeu]

x^(ln 2 + ln 2) + x^(ln 2 + ln 5) == x^(ln5+ln5) divide by x^(ln2+ln5): x^(ln2)/x^(ln5)+1==x^(ln5-ln2) substitute: t=x^(ln5)/x^(ln2) t^2-t-1==0 following steps are same

[@lirantwina923]

Please try to integrate 1/x^5+1

[@greeklighter-countryball68]

7:14 "How do we solve this equation though? Five over two raised to the lnx equals the golden ratio." "Yes"

[@mcalkis5771]

I believe you can tidy this up a bit by using the change of basis formula of the logarithm. As in: log(φ,5/2)=ln(φ)/ln(5/2) Thus, x=[exp(ln(φ))]^(1/ln(5/2)) x=φ^(1/ln(5/2)) Edit: Your solution might actually be prettier lol.

[@Fizban]

I was thinking of x=phi^(1/ln2.5). Is there a preferable method of writing it down? And if so, why?

[@Cynxcally]

I saw the thumbnail and tried solving it myself, And I got the answer (5/2)√[(1+√5)/2] as X.

[@dzbanekkulka7424]

Blackpenredpen just casually fit the golden ratio and 69 in one equation

[@MathwithMarker]

Nice equation❤

[@ironicanimewatcher]

i tried solving it on my own and got spooked by the golden ratio jumpscare

[@arunprasath.r2911]

Can you please help me with this question sir, x^2=y^2 by taking square root we have √x^2=√y^2 |x|=|y| what is the next step sir

[@tatlook]

I got (phi-1)^(1/(ln(2/5))), which is same.

[@LukeSeed]

That's a lot of pens you got there

[@tylerwebb7303]

if you told me when I clicked on this video that I'd be seeing a quadratic I would have told you that you were crazy

[@josepherhardt164]

If THIS doesn't smell like a hidden quadratic. Later Edit: And it was. And I haven't even seen the video yet.

[@davidsaioc2507]

If we want to find all the real solutions, we have to check negative numbers as well, but ln x has no real solution if x is negative, so using this way cannot give us negative solutions. My question: is there a way to prove that this equation has no negative solutions or is there a condition for x being positive? Thanks!

[@zactastic4k955]

The cool thing is phi is approximately 1.6 and the answer is 1.69 which makes since because e is approximately 5/2

[@error5487]

does anyone know the promo code (if there's any) for his merch?

[@m.h.6470]

Solution: x^ln4 + x^ln10 = x^ln25 x^(2ln2) + x^(ln2+ln5) = x^(2ln5) (x^ln2)² + x^ln2 * x^ln5 = (x^ln5)² |-(x^ln5)² (x^ln2)² + x^ln2 * x^ln5 - (x^ln5)² = 0 Substitution u = x^ln2 v = x^ln5 u² + vu - v² = 0 u = -v/2 ± √((v/2)² - (-v²)) u = -v/2 ± √(v²/4 + 4v²/4) u = -v/2 ± √(5v²/4) u = -v/2 ± √5 * v/2 u = v * (-1 ± √5)/2 Resubstitution x^ln2 = x^ln5 * (-1 ± √5)/2 |ln ln2 * lnx = ln5 * lnx + ln((-1 ± √5)/2) |-(ln5 * lnx) ln2 * lnx - ln5 * lnx = ln((-1 ± √5)/2) lnx * (ln2 - ln5) = ln((-1 ± √5)/2) |:(ln2 - ln5) lnx = ln((-1 ± √5)/2) / (ln2 - ln5) |e → e^(lna/b) = a^(1/b) x = ((-1 ± √5)/2)^(1/ (ln2 - ln5)) x₁ ≅ 1.69075... x₂ ≅ -0.567... + i * 0.167... Curious, that x = 0 doesn't come up as a result, even though it is clearly a valid solution

[@arthurvictor6704]

That's just brilliant! Literaly lol

[@radzelimohdramli4360]

Can x be equal to 1 as well?

[@TheFrewah]

No, 1 to the power of something is always one, you would get 1+1=1 which is false

[@gachanimestudios8348]

Whenever (•)² = (•) + 1 appears, always expect ±φ^(±1).

[@rualmenendez2421]

Im in 9th grade, and it's funny how i understood some of it. That's why because he explained in a way that my sped mind could even understand. And today, i thought myself the Pythagorean Theorem

[@metorasay]

3:40 doesnt that mean that 4+10=25?

[@API-Beast]

Wait, but why is the golden ratio (1 + sqrt(5)) / 2? that's such a random number!

[@abdsalam34567]

i love math of course

[@giuseppemalaguti435]

x=[(√5+1)/2]^(1/ln5/2)=1,69075

[@yogamulyadi2046]

x^ln⅖=-½(1±√5) x=-½(1±√5)^(1/x^ln⅖)

[@hamza201183]

Beautiful

[@scottleung9587]

Hey, I did it!

[@fubblitious]

Nice