Wow first time i saw exponential with a decimal number. Could you show it how we can calculate without a calculator.

-1^3.5 = -1^7/2 = √(-1^7) = √-1 = I

p + k = 41 pk = 400 k * (41-k) =400 k^2 - 41k + 400 +0 k = 1/2(41+/-(1681 -1600)^1/2) k = 1/2(41+/-9) k = 25, or 16

X=9 is correct.

P should be equal to 25 K should be equal to 16 Square root of p is 5 Squarecrootvof k is 4swuarecrootvof p + square root of k= 5+4=9 Square root of p multiples by squarecroot of k 25x16= 400 Square root of 400 is 20

Nice!!!

Nice

The equation is equivalent to 32 log_2 n=n. It is easy to see that n=2^8 is a solution of the equation. Since n'=1 and (32 log_2n)'<1 for n>2^8, the number 2^8 is the unique solution for n>2.

Wonderful

Good problem

This is so lame it's painful

WOW!!!

Kuwait Math Olympiad: x² - y² = 24, xy = 35; x, y =? x² > y² > 24; x² - y² = 24 = (12)(2) = (7 + 5)(7 - 5) = (± 7)² - (± 5)² = (± 5i) - (-/+ 7i) x = ± 7, y = ± 5 or x = ± 5i, y = -/+ 7i Answer check: x = ± 7, y = ± 5; x = ± 5i, y = -/+ 7i x² - y² = (± 7)² - (± 5)² = 24, (± 5i)² - (± 7i)² = 25i² - 49i² = - 25 + 49 = 24; Confirmed xy = (± 7)(± 5) = 35, (± 5i)(-/+ 7i) = - 35i² = - 35(- 1) = 35; Confirmed Final answer: x = 7, y = 5; x = - 7, y = - 5; Two pairs of imaginary value roots, x = 5i, y = - 7i or x = - 5i, y = 7i

great. Can you do some problems on negative angles?

Good

Good solution

could you not reject (-1-(5)^.5)/2 also because it is less than -1

Why did you make it so complicated? Simply square both sides, you will get your answer in 3 lines.

Not hard but nice

Let x^6 = 1. Six roots of unity. Then (2x)^6 = 2^6. Done.

Not bad

Nice!!!!

Again I did it

sin (theta) = cos ^ 2 ( theta) = 1 - sin^2 ( theta) (sin( theta) + 1/2)^2 = 5/4 sin(theta) =( √5 - 1)/2

Nice prob

5^(ln3/ln5)*5^(x^2)=5^(ln15/ln5) , x^2+x*ln3/ln5-ln15/ln5=0 , x=(-ln3/ln5+/-V((ln3/ln5)^2-4*ln15/ln5)/2 , x= 1 , -log(15)/log(5) test , x=1 , 3*5=15 , OK , test , x=-log(15)/log(5) , --> result 15 , OK , solu , x= 1 , -log(15)/log(5) ,

I did it

Just a little mistake at the end, the K2 solution is -2 and not -3.

You're a mathematical Magician.

Wonderful problem

Orang gila aja bisa. Ya k=2 ngga usah diurai2kan segala.

great again.

Sul

Excellent man!!

Nice solution

Only one answer 0 ?

Wonderful. You gave us suspension at m^2 -4m + 20 = 0 but brilliantly tackled.

m^2 + 4m + 20 = 0 was the equeation for other solution how you changed it to m^2 -4m + 20 = 0. People who solve like you will lose marks duw to carelessness. Its gonna have complex roots.

Good solution video

Nice solution

Welcome Backkkkk ❤

Easy question. Not expected in Olympiad

(a³)*

X = 9

Subscribe it's free 😄

(X+5)^X=7^2=(2+5)^2 X=2(A)

a=b=c= 0; +2^1/2 ;-2^1/2

3^x + 3^x + 3^x = 3^x (1+1+1) = 3^x.3^1 = 3^(x+1) and 9^9 = (3^2)^9 = 3^(2 times 9) = 3^18, so : 3^(x+1) = 3^18 x+1 = 18 x = 18-1 = 17, which is answer B