Germany - can you solve this exponential equation? find x!

[@boguslawszostak1784]

for x=0 KHS = 4^4 RHS = 0 so x0, and we can divide both side by x (x-4)^4/x^4=1 ((x-4)/x)^4=1 (x/4-1)^4=1 a= x/4-1 a^4=1 a= e^(n*pi/4) x=4*(a-1)

[@Psykolord1989]

Before watching: Looking at this visually, I can tell the real answer is 2 right off the bat; (-2)^4 = 2^4. There are also two complex solutions. To find those, we need to do a little work. Using the binomial theorem we get x^4-16x^3+96x^2-256x+256 = x^4. Subtracting x^4 from both sides yields -16x^3+96x^2- 256x+256 =0. We should do synthetic division, but first let's divide both sides by -16 so that the first coefficient is 1. That gives us x^3-6x^2+16x-16. Using synthetic division, we determine that X=2 is a factor. Thus, we can rewrite this as (x-2)(x^2-4x+8). We use the Quadratic Formula with the second factor. When we do so, the discriminant is negative. Thus, our other two roots are complex, not real. If looking only for real solutions, X=2 is the only solution. If accepting complex solutions though, the quadratic formula gives us (4±√-16)/2 = (4±4i)/2 = 2±2i Real solution: X=2 Complex solutions: x = 2±2i

[@renesperb]

A very simple way to find the solution: we have the possibilies x-4 = x,x -4 = -x ,x - 4 = i x ,x = - ix. Hence the 3 solutions are immediately clear.

[@prollysine]

we get , x^3 - 6x^2 + 16x - 16=0 , (x-2)(x^2-2x+8)=0 , x=2 , +1 -2 x^2-2x+8=0 , x= 1+i*V7 , 1-i*V7 , -4 +8 test , x=2 , (2-4)^4=16 , 2^4=16 , OK , +8 -16=0 ,

[@CharlesChen-el4ot]

(x-4)^4 - x^4=0 ((x-4)^2 - x^2)*(x-4)^2+x^2)=0 Sol 1. (x-4)^2-x^2 =0 -4*(2x-4)=0 x =2 Sol 2. (x-4)^2+x^2 =0 x^2-8x+16+ x^2=0 2* x^2 -8x+16=0 x^2 -4x + 8 =0 x = 1/2*(4 +/- (16-32)^1/2 x = 2 +/- 2 i

[@takacolonel4677]

日本の進学校の子なら簡単に解いちゃうね

[@masato3001]

それ、4?y?

[@すずけん-n8q]

a^4-b^4=(a^2+b^2)(a^2-b^2)=(a+bi)(a-bi)(a+b)(a-b)=0 => {a=-bi,a=bi,a=-b,a=b} => (x-4)^4-x^4=0 => {x-4=-xi,x-4=xi,x-4=-x,x-4=x} => { x=s+ti,s∈R,t∈R: {x-4=-xi => s+ti-4=-(s+ti)i=t-si => (s-t-4)+(t+s)i=0 => {s-t-4=0, t+s=0} => s=-t=2}, {x-4=xi => s+ti-4=(s+ti)i=-t+si => (s+t-4)+(t-s)i=0 => {s+t-4=0, t-s=0} => s=t=2}, {x-4=-x => x=2}, {x-4=x => No solution} } => {x=2-2i,x=2+2i,x=2}

[@hansolejennsen1539]

Ну и к чему все усложнять. Х=2.