Solving A Challenging Exponential Equation

[@anushkrajbordia1873]

The thing I like about this channel the most is this guy learns from his audience, which in my opinion shows humble nature & also his passion for mathematics is brilliant... Keep it up buddy! Love from India!

[@SyberMath]

I appreciate that! 💖

[@georgesbv1]

To factor out x-2, we divide by 100 and distribute the powers to simplify. Then apply logarithm 2 ^ (4(x-1)/x-2) * 5 ^ (x-2) = 1 2((x-2)/x)*log2 + (x-2)*log5 = 0 (x-2)(2/x*log2 + log5) = 0 (x-2)(xlog5 +2log2) = 0

[@SyberMath]

Wow! That's pretty!

[@beautifulmindinpuzzles7716]

Very clever method. I love it. Thanks for sharing.

[@Arijitians1]

Superb man👏👏👏excellent...

[@Muslim_011]

Finally someone share the same idea as mine

[@dileepmv7438]

@@Muslim_011 uh hu

[@SyberMath]

This is a really nice problem on exponential equations. One solution is kind of obvious but how do you find the other one? There are two methods as far as I know. Any thoughts?

[@leonhardeuler5211]

You can use Vieta’s formulas, namely the sum of the roots or the product. Say x_1=m and x_1*x_2=c/a then x_2=c/am

[@SyberMath]

That's very good!!!

[@mehmetdemir-lf2vm]

first solution is wrong. true solutions are x=-log(4)/log(5) and x=2.

[@micheleannoni8074]

@@mehmetdemir-lf2vm log4 is 2log2 and log 5 is 1-log2, hence the 'wrong' solution is equivalent to yours

[@mehmetdemir-lf2vm]

@@micheleannoni8074 i used mathematica to solve that. the first solution in the video is incorrect. see the solution (different from 14:51): www.wolframalpha.com/input/?i=Solve%5B16%5E%28%28x+-+1%29%2Fx%29*5%5Ex+%3D%3D+100%5D

[@voorth]

I would prefer to see the second solution simplified by cancelling 2: x = (2log2)/(log2 - 1)

[@SyberMath]

You're right. I forgot to simplify it.

[@MatteoBroggi]

Not only that... it can be written even more elegantly. Keep the minus in the second solution and use back the relation that 1-log(2) = log(5). Thus, you get -2log(2)/log5. Bring the -2 into the log in the numerator, you get log(1/4))/log(5). Finally, from the property of the logarithms that the ratio of logarithms of the same base is change in the base, you get x = log_5 (1/4). It looks much better written like this!

[@SyberMath]

@@MatteoBroggi I agree!

[@Arijitians1]

@@MatteoBroggi good

[@amtrakatsfnyc]

Another method is to use "100=25 times 4". This will generate an exponential equation in which one side has an exponent of (x-2) and the other side is "1". Allowing the claim that "x-2=0", hence "x=2". This is a 10 step [or less] solution. It avoids the need for logs.

[@CaradhrasAiguo49]

2:16 I wrote 16^[(x-1) / x] as 2^[4(x-1) / x - x + x], factored out the 2^(+x) to combine with the 5^x to form 2^[4(x-1) / x - x] * 10^x on the LHS to make the log base 10 slightly cleaner.

[@akramkhanafridi1755]

Dear Sir, Excellent Algebraic Skills. I love your way.

[@SyberMath]

Thank you! 💖

[@piyushdaga357]

Clever use of logarithms in substitution of log 2 as 1-log 5 ... Nice video...

[@SyberMath]

Glad you like it!

[@sekarganesan]

Agree...I tried to like twice it removed my previous like 😃

[@lsys6815]

Unbelievable!While watching your impressive solutions, my horizon gets a little wider. I wonder how many hours a day you worked to be like this one day. I'm a high school student and I'm watching your videos in desperation. Can you give me some advice that will save me, please. Tanks.

[@SyberMath]

Thank you! 💖 There are many people in my audience that know more math than I do. I just love solving problems and sharing my ideas. You need to start somewhere and practice! If you have a passion to learn math and apply your skills in problem solving, you will get to a higher level. It takes time, practice, dedication, patience, and a huge ambition! Take lots of notes. Review formulas and strategies. Most importantly, 💗 what you're doing! The rest will come... I hope this helps

[@onlyonlie891]

What mathematics should someone that barely graduated highschool prepare themselves for in regards to a basic 4 year term of college that isn't specializing in mathematics? And what should an aspiring physicist be getting familiar with?

[@aliasgharheidaritabar9128]

Best math channel for fun.

[@crackjeejeeadvanced1407]

It's my first video sir to c this good explanation and questions sir ILY

[@deepghosh7626]

Sum of the roots will be (2-4log2/1-log 2). One of the root is 2. We can use it to find other root

[@SyberMath]

Good thinking!

[@mehex9858]

I would like to solve this by the following method: 16^((x-1)/x) • 5ˣ = 100 2^((4x-4)/x) • 5ˣ = 10² 2^((4x-4)/x) • 5ˣ = (2×5)² 2^((4x-4)/x) • 5ˣ = 2² • 5² From here, on comparing both the sides you probably must have figured out that x=2 as you see that 5²= 5ˣ. But just to be sure enough you can also check the ones with base 2. 2^((4x-4)/x) = 2² (4x-4)/x = 2 4x-4 = 2x 4x-2x= 4 2x= 4 x= 2 So you got x=2 in both the cases. This means x is actually 2. :))

[@ahmet.t.g]

nice video, I like how you make/solve challenging questions that require a very different way of thinking

[@SyberMath]

Glad you think so!

[@sekarganesan]

Wow! I compared the powers of 2 and 5 to get x=2. Didn't expect another solution.

[@SyberMath]

Thanks!

[@justabunga1]

The other solution could have been simplified as 2log(2)/(log(2)-1) or log(4)/(log(2)-1)=log(4)/log(1/5)=-log(4)/log(5)=. If you want to expand more using properties of logarithms as a single logarithm, it would be -log base 5 of 4, or log base 5 of (1/4).

[@SyberMath]

Nice!

[@johnnath4137]

The sum of the roots is 4log2/(log2 - 1). If we know one of the roots is 2 then the other is 4log2/(log2 - 1) - 2 = 2log2/(log2 -1).

[@SyberMath]

That's good!

[@RexxSchneider]

Using log base 10 is not optimal for simplicity. After a little thought, it should be clear that we'll end up solving a quadratic and that the x^2 term in the quadratic will come from the 5^x term in the original equation. That suggests we take logs base 5. I'll use log to mean log base 5. We get: ((x-1)/x).log16 + x = log100 = log25 + log4 Writing a = log4 will make that easier to write, since log16 = 2.log4. Noting that log25 = 2, we get: 2a(x-1)/x + x = 2 + a 2a(x-1) + x^2 = 2x + ax x^2 + x(2a - a - 2) - 2a = 0 x^2 + x(a-2) - 2a = 0 That factorises to (x-2)(x+a) = 0 So x = 2 or x = -a. The negative solution for x is at negative log to base 5 of 4 which is about -0.861.

[@SyberMath]

Nice!

[@hookem7060]

I also use a quadratic equation: ax^2 + bx + c = 0; a = log(5)/log(16), b = 1 - log(100)/log(16), c = -1 Roots: x = 2, x= -0.8614

[@joaquingutierrez3072]

Nice video!! If you only want to find rational solution it is enough to write the prime factorizations in both sides and make exponents equal. That way get x = 2

[@SyberMath]

Thank you! That is good!

[@tonyhaddad1394]

You can use the 2 to factorise the quadratic equation (X-2)( ......) = ........

[@SyberMath]

That's right! Good finding.

[@chaosredefined3834]

You can do slightly better. The product of the two solutions is equal to c/a. So, 2*x = -4 * (log 2)/(1 - log 2). Or x = -2 (log 2)/(1 - log 2)

[@SyberMath]

@@chaosredefined3834 Absolutely!

[@RandomGuy-pe7zs]

Using the quadratic equation and the Sum and Product of roots, we can subtract 2 from (-b)/a in the quadratic eqn

[@ruiegao9724]

Possibly earlier to simplify the original to 4^(x-2) x 5^(x^2-2x) = 1 fist, then log both sides and then solve quadratic for the two solutions easily come out with x1=2, x2=-log5(4).

[@AncientDebris-1]

The answer is x=2 or log1/4 base 5

[@dsanghi100]

Can I write left and right side in powers of 2 and 5 and then just compare these powers to get value of x? 100 = 2^2 * 5^2. 16 ^((x-1)/x) = 2 ^ (4 *(x-1)/x). on the left side, you have 5^x and on the right side, you have 5^2 which gives x = 2. This satisfies powers of 2 as well. Since 2 and 5 are prime numbers, this should be possible.

[@SyberMath]

Sure!

[@davidseed2939]

The short cut here is to substitute 5=10/2 so 5^x=10^x.2^-x . Thus our equation becomes 2^4. 2^(-4/x).10^x.2^-x =10^2 For shorthand set log_10(2)=a ~=0.301 Take log base 10. 4a -4a/x + (1-a)x =2 (1-a)x +(4a-2) -4a/x=0 Multiply by x. (1-a) x^2 + (4a-2)x -4a =0 We can solve this using the quadratic formula, or by inspection of the original equation we spot the x=2 is a solution. So from Vieta the product of the roots is -4a/(1-a) so the roots are 2 and -2a/(1-a) ~= -0,602/0.699= mentally about -0.866

[@nacimyehia6841]

hi, why didn't you simplify 4.log2/ 2.log2 -2 to 2.log2/log2 -1 ,as both the nominator and the denominator were multiples of 2? Or am I missing anything? thanks in advance

[@iraklidiasamidze4400]

I got x=2 without the use of quadratic formula by setting the base of the logarithms to 2 (since the properties of log that we used work for log with any base, except obviously for base 1 or non-positive base), the equation then would be 2x - 4 = 0

[@clashingunitez4236]

16^x-1/x . 5^x = 100 4^2(x-1/x) . 5^x = 4.5² By equating co- efficient:- 4^2(x-1/x) = 4 2x-2 = x x=2 Also, 5^x = 5² x= 2 Edit - I know it's kids method .. but kids method are the most elegant to solve.

[@gabcalvert5856]

Very Elaborated ,but a nice piece of maths there.Keep it there.Great channel

[@SyberMath]

Glad you think so!

[@aashsyed1277]

sybermath, which app do u use to put these problems? or write these? which website?

[@mintusaren895]

It's equal to hundred of thousands years. And BRAVE MAN.

[@vishalmishra3046]

This is an exponents problem so take log on both sides and solve with algebra to get all the solutions. (1-1/x) 4 ln(2) + x ln(5) = 2 ln(10) = 2 (ln 2 + ln 5) => (2-4/x) ln 2 = (2-x) ln 5 => (x - 2) ln 4 - x (x-2) ln 5 = 0 => x = 2 or x = - ln 4/5 = log5 (.25). Substituting both 2 and log5 (.25) one by one, you can confirm both are correct solutions.

[@vyaspk9704]

Use Euclid's prime factorisation. I think that method will be easier.

[@SyberMath]

Thanks for the tip!

[@beastmode1647]

That would only find x = 2, the integer solution But the original problem never said that x had to be an integer. x = 2 log 2 / (log 2 - 1) is a totally valid, non-integer solution. Euclid’s prime factorization would miss this solution

[@em_zon2643]

Very clever use of math tricks! Good one!

[@hsshashidhargowda4349]

In what scenario we should consider log on both sides?

[@SyberMath]

when there's a product or a quotient

[@JohnRandomness105]

There is one obvious integer solution. But take the logarithm (natural, common, any base) of the equation, and one finds a second solution: x = (-2 ln 2)/(ln 5) -- again, any base will do. One gets a quadratic in x.

[@gustavinho1986]

The second solution can be rewritten as x=-2 log2/(1-log2)=-2 log2/log5=- log_5[4]=log_5[1/4].

[@佐藤広-c4p]

I've always wondered why when using the formula for the solution of the quadratic equation ax²+bx+c=0, if b is even (b=2b'), why another formula x=[-b'±√ {(b') ²-ac}]/a Isn't used? That is, in this problem, x=[(1-2log2)±√{(2log2-1) ²+4log2(1-log2)}]/(1-log2)={(1-2log2)±1}/(1-log2). Therefore, it can be calculated more easily as 2(1-log2)/(1-log2)=2 or -2log2/(1-log2)=2log2/(log2-1). If a is 1, the denominator disappears and it becomes easier. However, I have never seen a video using this formula. So my conclusion is that this formula is not taught in schools, that is, it is not official, so it is treated as "non-existent". It's a pity, I don't think it should not be used because of unofficial.

[@mangeshbhanage5444]

100 = 25 × 4 = 5^2 × 4 If we compare both sides Theres is only one term on both sides containing.and equality sign suggests that 16 ^ x-1/x = 4^1 X-1/x = 1 X=2...satisfies the equation

[@shivnarayanbakshi7463]

16^(x-1)/x • 5^x=100 => 2^[4(x-1)/x] • 5^x=2²×5² => 4(x-1)/x log2 + x log5=2 log2 + 2 log5 => Comparing both sides x=2 & 4(x-1)/x = 2 => x=2 [Answer]

[@Kkkkkkk-d8k]

Sir I am a great follower of you from India. Sir I have some great problems. How do I send it to you??

[@SyberMath]

Hi Joyita. Nice to meet you! You can follow me on twitter and then send them that way. Do you use twitter? Here is my page: twitter.com/SyberMath

[@Kkkkkkk-d8k]

@@SyberMath Sir actually this is Subhadeep Das. I am using my mom's phone. After I saw this message I opened up an account on twitter. I will be sending the problem over there. Thank you Sir!!

[@SyberMath]

Sounds good Subhadeep! You’ve been followed!

[@txikitofandango]

I solved it using the natural log, and that was simple enough, but your way is more elegant

[@SyberMath]

Thank you! 😊

[@XLatMaths]

If you do it as the quadratic ln5(x^2) + (ln0.16)x - ln16 = 0 you get x = 2 and x = log_25_(1/16) which I think is a cleaner answer.

[@ramaprasadghosh717]

A little bit of work leads to 4 ^((2(x-1)/x)-1) = 5 ^ ( 2-x) or 4 ^ ((x-2)/x)* 5^(x-2) = 1 or (4^(1/x)*5)^(x-2) = 1 or x=2 is a feasible solution

[@jeromemalenfant6622]

OR if you rewrite the second solution in terms of log 5 it simplifies to x-= 2- 2/log 5.

[@damiennortier8942]

Can you solve ABC = (A+B+C)^3 where ABC means 100A + 10B + C? Or with another number of unknowns or a different power please

[@devanshsharma2106]

if we know 2 is one root, we can use product of roots of quadratric eqn and directly find other root since both would be real solns😁

[@aashsyed1277]

best math channel

[@SyberMath]

Thank you!

[@CRSrikanthSymphonicResonance]

Oh genius It can be solved so simply Write 100 as 5^2 * 2^2 = 2^4(x-1/x) . 5^x One case x = 2 is evident when 5^x = 5^2 Second case 4x-4/x = 2 4x-2x = 4 X= 2 condition holds successfully

[@CRSrikanthSymphonicResonance]

Orlog can be applied here both sides 2log5 + 2 log2 = xlog5 + 4x-4/xlog2 Writer log10base 10 as log5 +log2 =1 write log5 as 1-log2 that’s it ..

[@sekarganesan]

Syber - I am assuming you solve problems on ipad. What app do you use I like the page breaks. Have been using jamboard looking for alternative

[@SyberMath]

Yes and I use Notability

[@JohnRandomness105]

Knowing that x=2 is one solution, factor it out in the quadratic equation. Perhaps synthetically divide by x-2.

[@pz3328]

I divided both sides by 100. I changed 100 in 4^1 * 5^2. Then I solved it : (x-1)/x=1 and x=2. So my only solve it's x=2, why? Is it possible to do it in that way, or i need to follow yours?

[@SyberMath]

There is another solution

[@SamsungJ-kk5nr]

Very good exercise .

[@SyberMath]

Glad you think so!

[@obneljeanlouis6469]

You are brilliant in maths man I learn so much watching your videos Please don't use the voilet color when writing , with a Black screen it's hard to read and follow through Thanks. Keep it up

[@SyberMath]

Thanks! I’ll look into it

[@crackjeejeeadvanced1407]

Can u pls solve mysterious questions of jee exam of India

[@mohdadlymohdarif8674]

i mean 16 power (x-1/ x) times 5 power (x) = 100 2 power (4x-4 /x) times 5 power (x) = 5 power (2) times 2 power 2 2 power (2x-4 /x) times 5 power (x-2) = 1 (2 power (2/x) times 5) power (x-2) = 1 so the power must be 0, x=2 Please correct me if i'm wrong

[@SyberMath]

Looks good to me

[@Arijitians1]

16^[(x-1)/x] × 5^x = 100 or, 2^[4(x-1)/x] × 5^x = 2^2 × 5^2 or, 4(x-1)/x = 2 and x = 2 or, (x-1)/x = 1/2 and x = 2 or, 2x-2 = x and x = 2 or, x = 2 and x = 2 Thus x = 2 is the solution..... Excellent work Syber👍👍👍👏keep giving us great videos....

[@terrymiller111]

What is the name of the "whiteboard" technology?

[@piyushdaga357]

Another answer is -2/log 5(of base 2 ) It can also be written as x=( -2*log2) /log 5 ... By the way, how can I send you my solution...

[@piyushdaga357]

In above case value of x is approximately -0. 861

[@piyushdaga357]

How can I send you my solution?

[@SyberMath]

Sure. If you want to send me an image, post it on twitter using the link twitter.com/intent/tweet?text=%40SyberMath%20Check%20this%20out! and then include that tweet link in the comments here. I'm hoping that you use twitter.

[@georgesbv1]

it is the same real solution, just simplified and applied the log2=1-log5

[@nicogehren6566]

great solution sir thank u

[@SyberMath]

Happy to help. Thanks

[@雅阿林]

log100=log4*25=log4+2log5, log can be changed into ln ,too

[@vjlaxmanan6965]

The quadratic in x can b set up very easily without all the extra manipulations :(

[@cdiesch7000]

very nice problem it is not necessary to specify the base of the log and simply factorize all the numbers to their 5s and 2s. knowing that 2 is a solution you get: (x-2)*(log(5)*x+2*log(2))=0 x=2 or x=-2*log(2)/log(5) which is independent of the chosen base

[@SyberMath]

Nice!

[@b.a.6731]

a very colorful video again :)

[@SyberMath]

I agree!

[@stanisawwojcik4483]

Ciekawe sposoby rozwiązań zadan

[@christopherellis2663]

X=2 16^½ × 5²=100 =5²×2² It's a simple division and multiplication solution

[@beautifulmindinpuzzles7716]

Nice problem. A good challenge for a grade 10 to 12 student. The decimal answer can be simplified a little bit more: (-log4/log5) Thanks for sharing.

[@SyberMath]

Thanks and you're welcome. Good point!

[@mrhatman675]

A nice fact to notice is that the function is an increasing function and it doesn t have a solution for x>2

[@davidseed2939]

at the last step dont get rid of the negative because the answer is negative x=-2log2/log5 ~ -0.86

[@siddharthmishra8510]

nicely done ....but i think there is no need of logarithm function ......you can simply do it by resolving every thing in power of x-2

[@SyberMath]

That's right! Thanks!

[@freddyfozzyfilms2688]

notice that (x-1)/x = 1 - 1/x. Thus, as x increases 1 - 1/x increases. So 16^{(x-1)/x} increases as x increases. Furthermore 5^x increases as x increases. Thus the left hand side of the equation is always increasing as x increases. So, there can only be one solution.

[@SyberMath]

Very interesting!

[@RexxSchneider]

But false because of a discontinuity at x=0. When x is negative and small, 16^(1-1/x) is very large and positive and 5^x is just less than 1. As x goes more negative 16^(1-1/x) rapidly gets smaller, asymptotic to 16, while 5^x slowly gets smaller, asymptotic to 0. It should be obvious that the product can take any positive value, including 100, which gives another solution. As it happens, we get that when 16^(1-1/x) = 400 and 5^x = 1/4. That is x = log(base5)(1/4).

[@CriticSimon]

I like this problem. Thank you.

[@SyberMath]

You're welcome!

[@ossaimoses5829]

I solved this problem by writing 100 as 4×25 and equate to same base

[@雅阿林]

In China ,we use lgX not logX to express log10 X,dfferent way

[@SyberMath]

Hmm

[@italixgaming915]

My solution (of course much faster): 100=2².5²=16^(1/2).5². Then our equation becomes: 16^((x-1)/x-1/2)=5^(2-x) 4.[(x-1)/x-1/2].log(2)=(2-x).log(5). Since x is not 0, we multiply by x: 4.[(x-1)-x/2].log(2)=x.(2-x).log(5). x².log(5)+2.x.(log(2)-log(5))-4.log(2)=0. We divide by log(5): x²+2x[log(2)/log(5)-1]-4.log(2)/log(5)=0. And now we conclude: we have a quadratic equation, we notice that 2 is solution, then we have two real solution and we know from the equation that their product is -4.log(2)/log(5). Therefore the two solution are 2 and -2.log(2)/log(5).

[@Ni999]

I started with 16^(-1/x) 5^x = (5/2)² 2^(-4/x) 5^x = (5/2)² -(4/x)log2 + xlog5 = 2(log5 - log2) x²log5 + 2x(log2 - log5) - 4log2 = 0 Etc I don't know, looked like less bookkeeping? I really appreciate your insight for changing log5 into 1-log2, very nice. 👍👍

[@crackjeejeeadvanced1407]

Legends use logarithms

[@ermattia]

Hey Syber, where are you from?

[@baptistebermond2082]

Why didn't you factor the polynomial by (x-2) to simplify everything?

[@SyberMath]

How do we factor it?

[@baptistebermond2082]

@@SyberMath Doing a polynomial division since x-2 is a factor since 2 is a solution of the polynomial. Slight question, are you a math teacher in real life or something like this?

[@SyberMath]

Yeah, something like that! :)

[@tonyennis1787]

"Guess and check" is a great way to lean the boundaries of a solution or otherwise gain some intuition about a function. It's a terrible approach to solving problems.

[@tarunmnair]

Nice, when i did it, i got the solutions as 2 and (2 - log4/log5). Then i used the log substitution as mentioned in the video log2 = 1 - log5, and i got solutions as 2 and (4 - 2/log5).

[@Godplayzdice]

100=25*4 = (5^2)*(16^1/2). X=2 is a solution. Logs will give other solution.

[@kurzackd]

sooo, has he shared the other way to solve this yet??

[@sandeeshsewanagala8829]

Superb 🔥

[@SyberMath]

Thanks 🤗

[@sifisomavimbela8838]

Too good 🔥

[@SyberMath]

Thank you!

[@database2517]

People thought einstein had some problem because he had an I.Q. higher than other people. Similarly mad people think I am fool. But the reality is that I am a genius.

[@الثورة-ص7ق]

After solving I get:(x-2)Ln(5×4^1/x)=0》x=0 or x=Ln(4)/Ln(1/5)

[@viny884]

*Can someone tell me this is correct?* 16^((x-1)/x)*5^x=100 16^((x-1)/x)*5^x=4*(5^2) 4^(2(x-1)/x)*5^x=4*(5^2) Dividing both sides by 4 and 5^x By Simplifying that we get: 4^((x-2)/x)=5^(2-x) Dividing both sides by 4^((x-2)/x) 1=(5^(2-x))/(4^((x-2)/x)) (5^0)/(4^0)=(5^(2-x))/(4^((x-2)/x)) Comparing numerator 2-x=0 x=2 Or Comparing denominator: (x-2)/x=0 x=2 or x=infinity

[@pruthviranir9374]

Thank you to solve in another way. But this can be solved easily by just equating the powers.

[@SyberMath]

Sure. ok

[@에스피-h8t]

Solution by insight 4×25=100 x=2

[@amitkkar9183]

It can be solved by using laws of indices,no need to use logarithm

[@shivalikachourishi364]

Thatswhy i believe that indians are so fast in mathematics

[@pulak96]

You never explain the whys of your problem-solving approach.

[@isimsoyisim9138]

100 =25*4 100=5^2 *16^1/2 x=2, (x-1)/2=1/2

[@mohdadlymohdarif8674]

i like your videos Creative solutions That's math

[@SyberMath]

Thanks!

[@vh73sy]

one integer solution x=2 one real solution x=-0.86135 & two complex solution right? 😉👍

[@GregWeidman]

Negative log base 5 of 4.

[@srimanthirupathi4159]

Instead of going in such a complicated way for the first solution (x=2) You could have simply done 4x(5sq) =100 or sq.root of 16 x 5 sq so x-1/x= 1/2 x=2 or simply 2=x in 5sq=5 raised to the power x

[@SyberMath]

Nice!

[@vcvartak7111]

If you omit red color would be better because it's not visible on the mobile screen

[@SyberMath]

You mean the pink that I used in this video starting at 8:08?

[@vcvartak7111]

Perhaps it's pink but on dark background it's very faint for me to read. May be my eyesight is not strong but other colors are visible

[@mapduch3933]

Math is already a hard language to follow for some people. I think you might have lost some people when you made an unnecessary substitution of log2 for U. I think you should keep it simple and write out the exact terms of the equation to make it easier to follow.