4^(3.5)=2^7=128>125=5^3. 4^(5311/4311) > 4^(3.5/3) >5

Amazimg video :))

Use log

x= 29

129

Kuch bhi samaj nahi aaya guru ji😢

2×(2000^2+111^2) =2×(4000000+12321) =8024642

111222=9×12358 12356 lies between 111^2 and 112^2 So m^2+m=m(m+1) lies between 333 and 334 So m=333 or -334. E.g. 64<72<81,72=8×9 or (-9)(-8) m(m+1) lies between m^2,(m+1)^2

రె

You should have put the answer back into the expression to check it, I'm not sure it works

(x ➖ 3x+2). (x ➖ 3x+2) 5^4311<4^5311

This video made me realize that you can chain powers of x as long as you want, if eventually you set it to the power of a and let that result in a, then x = a^(1/a) is always a solution

pretty cool video

Nice Exponent Math Simplification: [(2² + 11² + 19)¹⸍²]³ = ? 2² + 11² + 19 = 4 + 121 + 19 = 144 = 12² [(2² + 11² + 19)¹⸍²]³ = (12²)³⸍² = 12³ = 144(10 + 2) =1440 + 288 = 1728

PLEASE TAKE AN EASY ROUTE.

Not equal to 0 because 0 is not +ve no,

Ans=a=2. b=3. & c= 4. (proof "2^2=4. ,2^3=8 &. 2^4=16. Now. 16+8+4=28.

a+b+c=2+3+4=9

I found it easier to evaluate **with** a calculator, haha haha

X=11/24

In this way the question was solved in 3 steps with same right answer

I took 9 common from whole expression and kept taking until reached the last like this: 9(1+9(1+9(1+9(1+9))))

a not equal 0 because 28 is pair if a = 1 then 2^a+2^b =26 26 = 2 x 13 no good then a =2 24 = 2 x 2 x 2 x 3 = 2^3 +16 then b = 3 16 = 2⁴ then c = 4 then

(x^3)^2= x^9 (28)^2= 784 {784 ➖x^ 9}= 775 7^10 3^25 7^2^5 3^5^5 1^2^1 3^1^1 23 (x ➖ 3x+2).

Very engaging problem

thankyou very much, i just learned something new

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.125 = 1/8 = 4^(-1.5) [since sqrt4 = 2, then 1/8 = 1/4 * 1/2, where 1/4 = 4^-1 and 1/2 = 1/(sqrt4) = 4^-.5] x-2 = -1.5 x = .5

Cool video bro. The explanation is understandable for any people watching. Keep up the good work. 👍👍

Great solution, and nice usage of the identity 3^3 + 4^3 + 5^3 = 6^3. However, the last step kind of irritates in terms of complex analysis, as there are 3 solutions for x. Those solutions are: 606, 606*e^(2i*pi/3) and 606*e^(-2i*pi/3). The general form of the solutions is 606*e^(5ni*pi/6), when n is an integer. Those 3 solutions form a circle when put in the complex plane, which is logical following the e^ix identity. Definitely a topic worth reading about! If anyone has any questions, I really love the subject so will be happy to answer

Solve given Exponential Equation: x³ - 303³ = 404³ + 505³; x = ? x³ = 303³ + 404³ + 505³ = (101³)(3³ + 4³ + 5³) = (101³)(27 + 64 + 125) = (101³)(216) = (101³)(6³) = 606³; x = 606 Answer check: x³ - 303³ = 404³ + 505³; Confirmed as shown Final answer: x = 606

2^a + 2^b + 2^c = 28 = 4(7) = 4(1 + 2 + 4) = 4 + 8 + 16 = 2^2 + 2^3 + 2^4 a = 2 < b = 3 < c = 4

(a + b)^2 = 8ab (a - b)^2 = 4ab [(a + b)/(a - b)]^2 = 2 (a + b)/(a - b) = √2

Nice trick

Math Olympiad: If, 444/(x + 555) = 666; 777/(x + 666) = ? x + 555 = 444/666 = 2/3, x = 2/3 - 555, x + 666 = 2/3 - 555 + 666 = 2/3 + 111 777/(x + 666) = 777/(2/3 + 111) = 3(777)/[2 + 3(111)] = 2331/335

I got all these steps but I was waiting for a trick for the last step. No trick there

Kis class ka hai ye sawal Maine kar liye maths enthusiasts hu . B tech 2nd year

Simpler: You've got a Pythagorean triple. 5^2 - 4^2 = 3^2 So K*5^2 - K*4^2 = K*3^2 Let K= 1111^2 and combine the squared product terms. 5555^2 - 4444^2 = 3333^2

Simpler: You've got a Pythagorean triple. 5^2 - 4^2 = 3^2 So K*5^2 - K*4^2 = K*3^2 Let K= 1111^2 and combine the squared product terms. 5555^2 - 4444^2 = 3333^2 [Edit: Basically what Leo said below.]

😂😂

It's better to write (1111*5)²-(1111*4)² Which will lead you to 1111²(25-16) 1111²(9) Then rewrite 9 as 3² as commonly applied on Pythagorean theorem applications 1111²(3²) And re write the expression 3333² Which is equal to the solution provided in the video, but it comes to show a beautiful expression

Easy 1111^-2 🧐

Excellent 👌

My dumbass would have just brute forced the squares and add them together.

(1111)(9999) = (1111)(10000 - 1) = 11110000 - 1111 = 11108889

You could also simply each k in the expression, eventually reaching k^(7/8)

(9/4)^9/4 (3/2)^9/2 (3/2)^9 × (3/2)^1/2 (19683/512) ×(√3/2) (19683/512) × (√3/2) (6561 ×3 /512) ×(√3/2) (81^2 ×3 /512)×(√3/2) (81^2×3/256×2)×(√3/2) (81^2×3/16^2×2)×(√3/2) (5.0625^2 ×3/2)×(√3/2) im tired ill complete this later

we can also break the exponent into 3/2×3/2...and then do 2 successive breakdowns?

16+8+4 pretty obvious

Nice video 👌