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The empathy and compassion shown here are a testament to the strength of the human spirit.🍭

Teacher please make a video on unit digit .. Such as... a^b .. if we divide this type of number ... what would be unit digit .

If you love math ,then you deserves this like button (BTW 1st comment)🗿 👇

could it also be done as 3logx base x = x. Then log will cancel leaving x = 3

This one was easy but fun. That's the magic of Mathematics.

Clean!!

Not often I get the right answer before watching the video. Living and learning!

Well obviously 3, obviously 1,... Third one is -1 im assuming

He is like the Bob Ross of painting, I swear that voice of his is so calming, plus he's soooo good at teaching maths, I wish I wasn't prepping for exams that just focus on you to learn as much as possible but rather be taught on what you wish to learn from him

Will done ❤

answer is 16, a power tower of 2's stacked 3 high. Graham's number (G64) has its origins from a massive power tower of 3's i believe.

We can have 9 to the power of the first equation, 16 to the power of the second equation, and 25 to the power of the third equation. It’s an alternative solution and may be simpler to understand.

you used a logarithm to make x^(x-3)=1 a logarithmic ecuation, and the argument of a logarithm is DEFINED as positive, so you should have restricted x to x^(x-3)>0, that's it a simple explanation, then , you can say, x^(x-3)<0, so you multiply everything for -1 and THEN convert it to a logarithmic ecuation.

I love that our video are easy enough that I can always work out the problem myself first, but not so simple that it feels trivial. As for my solution to this problem, it started out the same (excet I didn't show (1,1,0) was a solution until the end) up to the point where you show that x^2-2x+1+z^2=0. Originally I was going to use the quadratic formula like you did, but I noticed that it's just a sum of squares: (x-1)^2+z^2=0. Because a real number can't square to give a negative, both (x-1) and z have to be 0. The solution is then obvious from there.

Yo this is so cool! My friend did the Euclid contest and he gave me his sheet of problems and I was actually able to solve this particular problem, small world!

great material!

amazing proof

This saved me big time, I passed my Calculus 1 exam. Thank you Mr Prime Newton. Not only this videos but , the rest too🥺

Cant you just multiply the exponents : 10^10^10 = 10^10•10 = 10^100 not 10^ten billion

dovrebbe esserci un'altra soluzione reale con xyz = -6

exactly the way i did it !

Thanks mister ❤

16=2^4

3^2 doesn't have a double zero after the last whole number

great video! I had trouble figuring out how to sneak the high power into the root! and you helped me a lot

First time hearing of this 16 from 2^2^2

Bro you cleared all my doubt

You are a good mathematician .Keep it up !

I think the "trick" is to recognize that multiplying them all together leads to a nice equation where both sides are 4th powers, and then you can quickly make progress. I spent a lot of time trying to add and subtract and factor before watching the video, and it just took so long to get anywhere. But if you multiply everything it simplifies nicely.

I love your channel, do you like physics.?

We can also take take x³ common as in num it would be like x^6/2 2 bcz of square root thing and sqrt (9 - 1/x⁵) which will convert into - sqrt 9=3

16 😊 plz don't bother to heart, i have seen comments with 16 and hearts i know mine is also 💯. Thanks for this video.

I love this guy's charisma

Forbidden exponential

best zinger ever

😊👍

Excellent Sir.

Love your videos, you are really good at explaining!

Very helpful

You are the best teacher of this year

It's 16

What is the order rule lLATE

Calculate the number of #2 pencils required to write down the answer. Show your work. It's due in the morning.

Step 1: Make the base the same for each equation: log9 x + log9 y + log9 z² =2 log16 x + log16 y² + log16 z = 1 log25 x² + log25 y + log25 z = 0 Step 2: Combine logarithms using log x + log y = log xy: log9 xyz² = 2 log16 xy²z = 1 log25 x²yz = 0 Step 3: Remove logarithms using logb x = y => x = b^y: xyz² = 9^2 = 81..................(1) xy²z = 16^1 = 16................(2) x²yz = 25^0 = 1...................(3) Step 4: solve for y using the equation (3): y = 1/(x²z)..............(4) Step 5: Substitute y = 1/(x²z) into equation (1) and solve for z: z = 81x................(5) Step 6: Substitute this to equation (4): y = 1/(81x³).............(6) Step 7: Substitute y = 1/(81x³) and z = 81x to equation (2): x * 1/(81x³)² * 81x = 16 => x^4 = 1/(81*16) => x = 1/(3*2) = 1/6. Solution x = -1/6 won't do. All x, y and z must be positive because otherwise the logarithms aren't defined. Step 8: Calculate y and z using (5) and (6): x = 1/6 y = 8/3 z = 27/2 Checking these with the second equations indicates correct values.

I remember this question from when i took this test lol 😅

yoo i actualy joined this contest this year, i remember this question lol, this is one of the 5/10 questions that i can do

x = 1/6 y = 8/3 z = 27/2

Sachin sir from pw also give this que in class

With proper factoring, you get the linear equation Ma=b where M is the matrix [[1 1 2][1 2 1][2 1 1]] (all ones except the off diagonal), 'a' is the vector [log(x) log(y) log(z)], and b is the vector [log(81) log(16) log(1)]. Computing the inverse matrix and solving for 'a' gives you that a=[log(1/6) log(8/3) log(27/2)], thus the arguments of the logs in 'a' are the values for x,y, and z.

Fantastic video! Very clean solution