(k ➖ 7k+7).

8 devide 27 = 0,2962 equal m has 0,6639 those degreeded 3 has 0,6639 x 0,6639 x 0,6639 equal quiteful as reach 0,2962 fixed. Regards from indonesian javanist tolerance culture

a^2/49=7 (a ➖ 7a+7) .49/a^2=7(a ➖ 7a+7).

{0+0 ➖} ^1=1^1 (x ➖ 1x+1).

2^4^2^3 ➖ 1 1^1^1^2^1 ➖ 1 2 ➖ 1 (x ➖ 2x+1).

32^33 1^2^3^1 2^3 (m ➖ 3m+2).

(10^10) (2^5^2^5) (1^1^2^1) (2^1) (x ➖ 2x+1).

{108+108}=216 108^108 10^10^8^10^10^8 5^5^5^5^2^3^5^5^5^5^2^3 2^3^2^3^2^3^2^3^1^1^1^2^3^2^3^2^3^2^3^1^1^1 1^1^1^1^1^1^1^1^1^1^1^1^1^1^2^3 2^3 (K➖ 3L+2).

(k ➖ 2k+2).

5^10^5^10vs7^7^3^17 5^2^5^5^2^5vs1^1^3^17^1 1^1^1^1^2^1vs3^1^1 2^1v<3^1 (n ➖ 2n+1) (a ➖ 3a+1). 50^50<49^51.

5^10^5^10vs7^7^3^17 5^2^5^5^2^5vs1^1^3^17^1 1^1^1^1^2^1vs3^1^1 2^1v<3^1 (a ➖ 2a+1). (b ➖ 3b+1). 50^50<49^51.

6^6 2^3^2^3 2^1^1^3 23 (n ➖ 3n+2).

8^(10) 8^(2^5) 2^3^(1^1) 2^1(x ➖ 2x+1).

8/3^3 2^3/3^1 2^1/3^1 2/3 (m ➖ 3m+2).

3^3 (x ➖ 3x+3) .

is your way of mutiplying faster? the one i learned from school is a whole lot easier

Man just triple root to both

m³=64/27 m³=(4/3)³=(4/3)³*e^(2*k*π*i) m=4/3*e^(2*k*π*i/3) {k=0, 1, 2} k=0; m1=4/3 k=1; m2=4/3*e^(2*π*i/3)=4/3*(-1/2+√3/2*i)=2/3*(-1+√3*i) k=2; m3=4/3*e^(4*π*i/3)=4/3*(-1/2-√3/2*i)=2/3*(-1-√3*i)

let me guess m=1/2 is the only real solution.

You only have three solutions, because 3+-√3i/6 is the same as 1+-√3i/2 if you devide it by 3

0.01^-0.5207=11.000185275062208727304006065456 - ploss🤣

(0.01)^x=11 100^-x=11 x=-log11/log100 x=-log11/2 x=-0.52069634257911252037509998562151 0.01^-0.52069634257911252037509998562151=11

3^3 > 2^4 4^4 < 3^5 50^50 > 4^4 49^51 is obviously bigger than 50^50 but how much? 49^51/50^50= 49*49^50/50^50= 49*(49/50)^50= 49*0.98^50= 17.844314324268736195651325091206 times bigger

27 😊😊

Amazing

There is only one solution !!!! X=6 - 4 * sqrt 2 Solution X=6 + 4 * sqrt 2 FALSE!!! First subtract after quadrat like the first comment, otherwise you generate the false result.

Calculation in the video is complicated and in addition x1 (first solution) is not correct. If we put x1 to the equation √2 + √x = 2 it does not fit.

Thanks

There's a general rule for this: a^b - a^(b-1) = a^(b-1)(a-1). Take a simple example: 10^3 - 10^2 = (10^2) * (10-1) = 1000 - 100 = 900.

Even if there are alternative procedures to solve this equation, this is still a good review of an area of algebra that I studied some time (some LONG time!) ago. Thanks for posting this.

Sir very good question ❤❤❤❤❤❤❤

i suggest to subtract root 2 on both sides

X=1 is The Only solution. You can't take log from complex number!

3(n^1/2)=2 n=(2/3)^2=4/9

Much easier is to leave square roorh x on 1 side and then square both side. I got x1 only, not x2.

8*9^6=8*81^3=648*6 561=4 251 528

Can't we do sum like this 3^3x = 27 We can write 27 as 3^3 3^3x = 3^3 The bases are same so We will get this equation 3x = 3 x=3/3 x=1 1 satisfy the equation Check 3^x = 27 Substitute the x value 3^3(1) = 27 3^3 = 27 27 = 27 L.H.S = R.H.S IS IT RIGHT SIR?

3^4^5+3^4^2 3^2^2^2^3+3^2^2^2 1^1^1^1^1+3^1^1^2 3^2 (x ➖ 3x+2).

{0+0 ➖} ^1=1^1 (x ➖1 x+1).

2^5^2^5vs9^11^1 1^1^2^1vs^3^2^1^1 (x ➖ 2x+1) .2^1vs3^1(x ➖ 3x+1).2^1<3^1. 10^10<9^11.

{2+2 ➖ }+n^2+{n+n ➖ }={4+n^2+n^2}=4n^4 2^2n^2^2 2^1n^1^1 2n^1(n ➖ 2n+1).

{3+3 ➖ }+x^2+{x+x ➖ }={6+x^2+x^2}=6x^4 3^3x^2^2 3^1x1^2 3x^2 (x ➖ 3x+2).

{4+x^2}=4x^2 2^2x^2 1^1x^2 1x^2 (x ➖ 2x+1).

6^{n+n ➖ }+{7+7 ➖ }=6^{n^2+14}=6^14n^2=84n^2 6^7^7n^2 6^3^4^3^4n^2 3^3^1^2^2^1^2^2n^2 3^1^1^1^1^1n^2 3n^2 ( n ➖ 3n+2). 7^{n+n ➖ }+{6+6 ➖}7^{n^2+12 }=7^12n^2 84n^2 7^6^6n^2 3^4^3^3^3^3n^2 1^2^2^1^1^3^1n^1 2^1^3^1^n 2^3^n (n ➖ 3n+2).

2^(5) 2^(1) (m ➖ 2m+1). m^2^16 m^2^4^ m^2^2^2 m^1^1^2 m^1^2(m ➖ 2m+1).

(k ➖ 3k+3). 27 3^3 (k ➖ 3k+3).

{10+10}=20 10^10 5^5^5^5 2^3^2^3^2^3^2^3 1^1^1^1^1^1^2^3 2^3 (n ➖ 3n+2).

Go to 3:00 to skip useless filler. Nor sure why all these mathematicians are so verbose. LOL The thing is, anyone who watches one of these already knows about commutative, associative, exponents, logs, algebra, substitution, etc and do not require excruciating details. Just a suggestion.

K=3

2√n+√n=2 (2√n+√n)²=2² (2√n+√n)*(2√n+√n)=4 2√n*2√n*+2√n*√n+√n*2√n+√n*√n=4 4n+2n+2n+n=4 9n=4 n=4/9 🙂