Why do we have to subtract

Why do we have to subtract

Note that if you draw all solutions of f(x) = x^12 - 1 as points in a complex plain, the result will be an analog clock.

When you cube root both sides at 3:51 there is only one positive value. not + or -

EXCELLENT !!!!

Thank you

Thank you, a very clear explanation.

Please explain how you knew to factor -150 into -125 - 25. I mean, in this case, it's kind of obvious, but what about the general case where the numbers are not so convenient (i.e., a prime)?

Thanks

EXCELLENT !!!!

Comments is also 11😂😂

1 3÷ P 10

Thanks helped a lot

Jak pastuch krowie !...mogłeś od razu logarytmować, dużo zbędnej pisaniny.

Much easier method (when x E R): 2^x + 8^x = 68 2^x + 2 ^3x = 4 + 64 2^x + 2 ^3x = 2^2 + 2 ^6 (the same base and the same operation (adding) we can comparise all exponants) x = 2 and 3x = 6 --->x = 6/3 ----> x = 2

THANKS PROFESOR !!!

this is not the right way to do math

4 unknown exponent=64

I know this is an example and more complicated problems wouldnt be nearly as easy to "guess" the answer, but i realized x=2 just by looking at original question. However, learning how to re-write and manipulate equations is very useful. Thank you for the video

x=-1, 3/2

8

You should put in factor those identical.

This is easy

In the 2nd case , it has to be ( i . square root 3 ) and not square root of 3i .

this is good video

Nice

X=0.

EXCELLENT !!!!

x1=0; x2=1/(1-ln3/ln2)≈-1.71

X= (log64/log0.25)+1=-2.

Sorry don't understand

Agonizingly slow!!! Almost 10 minutes for what should be about 1 minute.

🇺🇸🇩🇴🇺🇸🇩🇴

Good job!

You reject x^2 = -9 at about 6:40 but at the start you do not restrict the solution set to Real numbers. Therefore Complex numbers need to be included as possible solutions.

Why such a long time for a simple one ?

Know your powers of numbers. 3: 1, 3, 9, 27, 81 2: 1, 2, 4, 8, 16 3^n > 65 81 = 3 ^4 81 - 65 = 16 = 2^4 Therefore n =4 it is that simple.

Is this working correct i dont think

n=4 is the answer

Wow! Thank you very much, you give me real challenging question i did not know where to even start

I was able to do this mentally without any paper. You are extremely slow and we had to watch this 2x speed and still slow. You need to speed up my brother and raise your level to Asian

Thanks 😊

"Faster than a calculator"

software engineer here: 2^16 = 65536. Just multiply by 4 and substract 1.

Solution by Euler's equation. Draw a circle of radius 10 centered on the origin of the complex plane. The four roots are where the circle intersects the positive real axis, the positive imaginary axis, the negative real axis, and the negative imaginary axis. Thus, x_0=10, x_1=10i, x_2=-10, and x_3=-10i.

A lot of this was useless and unnecessary Example : it was not necessary to rewrite the problem at the beginning when it was already written

How does this trash have more views than my video on the Sowwyfututut theorem

nice vid

❤❤

The answer is correct but the method is wrong! Many youtubers also make the same mistake. You will realize your mistake when applying your method to the problem x^u - y^t = 17. You will never find the following solution of the problem 5^2-2^3=17. You also cannot find the a simple solution of the above equation is 18^1-1^18=17.The correct way to do it is as follows: We realize that x and y cannot both be even numbers or odd numbers. - If x is even number and y is odd number, we find the a simple solution of the above equation is 18^1-1^18=17 with x=18, y=1; In case x,y >= 2 => x^y≡0 (mod 4) and y^x≡17≡1 (mod 4) => y^x+17≡2(mod 4)=>x^y - y^x ≠ 17. - So the following case remains, x is odd number and y is even number => we can easily find the fist solution x=3, y=4. We will prove this is the only solution. * First, we will prove that x must be smaller than y. By examining the function f(t)=ln(t)/t, we see that this function is inverse when t>=3 => If x^y = y^x+17 > y^x then x<y. * Then, we will prove that 3^4 - 4^3 = 17 < x^y -y^x for all numbers x, y such that y>x>3 => x= 3+a and y = 4+b with b >= a >= 1 => P = x^y - y^x = (3+a)^(4+b) - (4+b)^(3+a) = (3+a).(3+a)^(3+b) - [1+(3+b)]^(3+a). Because 3+a =< 3+b => (3+a)^(3+b) >= (3+b)^(3+a) => P >= (3+a)(3+b)^(3+a) - [1+(3+b)]^(3+a) = [(3+b)^(3+a)].{(3+a) - [(1+3+b)/(3+b)]^(3+a)} = [(3+b)^(3+a)].{(3+a) - [(1+1/(3+b)]^(3+a)}. And because 3+a<=3+b and [1+ 1/(3+b)] > 1 => [(1+ 1/(3+b)]^(3+a) < [(1+ 1/(3+b)]^(3+b) < e (e=2,71828….). => P > [(3+b)^(3+a)].{(3+a) - e}. And [(3+b)^(3+a)] >= 4^4 = 256 => P > 256.(4-e) > 17. Q.E.D