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0 with 3 complex solutions

1.2

X= (-4, 0, -2+2i, -2-2i)

2^12 = 4^6. (2+x)^6 = 4^6 x + 2 = 4. So x = 2. Solved in 3 seconds.

(25/4)^(9/4) = (25/4)^(2) x (25/4)^(1/4) = (25)(25)[√(5/2)]/16 = (25)(25)[√5)/(16√2) = (25)(25)(√10)/32

X= -4

No need to resolve such the above-mentioned way since it can be easily solved as x=1, isn'i it? Why made simple things complicated?

X= 1

(9^-2^-1)^-3 1/(9^-2^-1)^3 1/(1/(9^-2))^3 1/(1/(1/(9^2)))^3 1/(1/(1/(9^2)^3)) 1/(1/(1/(81)^3)) 1/1/1/(81)^3 1/1/1/(81)^3 ...lost

this problem can be solved in your head in 2 minutes. Your solution reminded me of a joke about a gynecologist changing piston rings in a car through the exhaust pipe.

Why do you complicate so much! X^5 -1 =0 X^5 = 1 Square root of 5 both side Then X = 1 You may prove it

Hcf 2⁹⁹(2-1) 2⁹⁹(1) 2⁹⁹

X=1

3

1.5

(X+2)^6=2^6×2=(2^2)^6=4^6 X+2=4 X=4-2=2

C=root3.......maybe

(X+2)⁶ = 2¹² multiply both side by 1/6 exponentially = x+2 = 2² x+2 = 4 x = 2 period!

১. ৬৭৭৭৭৭৭৭৭৭৭......

If you take both sides to the 1/6 power, you get x+2=2^2; x=2. Thereafter, you could expand the expression on the left using Pascal’s triangle, and divide using synthetic division to get the other roots.

We could reject the complex answers. Simply make it (x + 2) = +/- 4. So x = 2, -6.

normal

There's a saying KISS ( Keep It Simple & Stupid), but your way is Keep It Silly & Stupid.

Use 1^x to denote exp(2πix) = cos 2πx + i sin 2πx. Lemma: the solutions to x⁶ = 1 are x = 1^{n/6} for n = 0, 1, 2, 3, 4, 5. Proof: at most 6 solutions to a 6th degree polynomial equation can exist. Each of the proposed solutions checks out (just multiply their exponents by 6), and there are 6 of them. Therefore they are all of the solutions. QED. (x + 2)⁶ = 2¹² (= 4⁶) if and only if ((x + 2)/4)⁶ = 1 if and only if (x + 2)/4 = 1^{n/6} for n = 0, 1, 2, 3, 4, 5 (by the lemma), if and only if x = -2 + 4·1^{n/6}, for n = 0, 1, 2, 3, 4, 5. Those are your 6 solutions. In more prosaic and familiar terms: x = -2 + 4(cos πn/3 + i sin πn/3), for n = 0, 1, 2, 3, 4, 5. By the way: x⁶ - 1 = (x - 1^0) (x - 1^⅙) (x - 1^⅓) (x - 1^½) (x - 1^⅔) (x - 1^⅚) = C₁(x) C₂(x) C₃(x) C₆(x), where: C₁(x) = (x - 1^0) = x - 1, C₂(x) = (x - 1^½) = x + 1, C₃(x) = (x - 1^⅓) (x - 1^⅔) = x² + x + 1, C₆(x) = (x - 1^⅙) (x - 1^⅚) = x² - x + 1.

Hey, how did you get 2.5? You’ve lost me there.😮

Final answer is x= 1+ log3/2log i

(x+2)⁶=2¹² (X+2)⁶=4⁶ taking 6th on both sides 6√(x+2)⁶=6√4⁶ a√c^b=c^b÷a (X+2)^(6÷6)=4^(6÷6) X+2=4 X=4-2 X=2

x_j=4[cos(jπ/3)+i*sin(jπ/3)]-2, j=0,1,...,5 & i=√(-1)

(x+2)^6=2^12=(2^2)^6=4^6 [{(x+2)/4}^2]^3=1 {(x+2)/4}^2=1 or, w or, w^2= 1 or, w^2 or, (w^3.w) = 1 or, ( w^2 ) or, ( w^4 ).........( where w is the complex cubic root of 1, w=(-1+i(sqrt(3)))/2,,,w^3=1 (x+2)/4 = ( + / - ) 1 or, ( + / - ) w or, ( + / - ) w^2 then , the solution , x = - 2 ( + / - ) 4 or , x = - 2 ( + / - ) 4w or, x = - 2 ( + / - ) 4w^2

X=2

X=0

what a strange stupid non realistic equation. If you draw a graph of each equation on paper, the X and Y never meet as the solution of a real quadratic equation 😮

[(x+2)³]² - [(2⁶)]² = 0 ((x+2)³ - 2⁶)((x+2)³ + 2⁶)) = 0 ...so it's a matter of solving these 2 qubic equations: 1) (x+2)³ - (2²)³ = 0 2) (x+2)³ + (2²)³ = 0 1) (x+2)³ - 4³ = 0 [(x+2)-4]([x+2]² + 4² + 4[x+2]) = 0 [x-2](x²+4+4x+ 4² + 4[x+2]) = 0 [x-2](x²+4+4x+ 4² + 4x+8) = 0 (x-2)(x² + 8x + 28) = 0 ...this is TEDIOUS but hey I surely can solve it - if you pay me 😂

Essay

×=2

Ne siktirik bir soru

√22801=151// x^2=22801 x=151//

Log 8 32

2

This question is answer oriented, if the answers are different then we can not solve this with this solution probably 🤔

X=5/4

x^x=2^2x Методом подбора: Если x=0, тогда 0^0=2^2*0 Потому что оба выражения верны одному и тому же значению: 1 2*0=0, а ноль это показатель степени, а любое выражение в нулевой степени равно одному и тому же значению - единице.

Quicker to do the multiplications from scratch although more likely to make an error that way

This guy is an idiot,

too long to solve, mine is much shorter. x power of 5 = 1 then square root of 5 both sides x = square root of 5 of 1 then x = 1

x=log200=log100+log2=2+log2 x≈2+0.30103≈2.30103 3×10^2.30103≈600.000006

As an America High School Math teacher this problem is a simple case of following math rules. It is clearly 27 like he states. There is no ambiguity.

개소리를 정성껏하네 ㅋㅋㅋㅋㅋㅋ x=1

I've got 3+log 4 ⁵

Lesson 42: How to make a short story long