We could subtract (x-4)^(x+3) to both sides, which we get (x-4)^(x+5)-(x-4)^(x+3)=0. Factor out (x-4)^(x+3), which is (x-4)^(x+3)((x-4)^2-1)=0. Set each factor equal to 0, so (x-4)^(x+3)=0 and (x-4)^2-1=0. (x-4)^(x+3)=0 has one solution, which is x=4. (x-4)^2=1 making this x-4=±1, xo x=3 and x=5. We have all 3 solutions, so therefore, x=3, 4, and 5. If you graph those two solutions and using the calculus method, we only see that there are two intersection points at x=4 and x=5 since x=3 is out of the domain. Therefore, the domain for those two functions has to be x≥4.
I set y = x - 4 and moved the entire equation to one side to get: y^(y + 9) - y^(y + 7) = 0 y^(y + 7) (y^2 - 1) = 0 Therefore: y^(y+7) = 0 or y^2 = 1 This gives me y = -1, 0 or +1. Going back to x by adding 4 gives: x = (3, 4, 5)
We could subtract (x-4)^(x+3) to both sides, which we get (x-4)^(x+5)-(x-4)^(x+3)=0. Factor out (x-4)^(x+3), which is (x-4)^(x+3)((x-4)^2-1)=0. Set each factor equal to 0, so (x-4)^(x+3)=0 and (x-4)^2-1=0. (x-4)^(x+3)=0 has one solution, which is x=4. (x-4)^2=1 making this x-4=±1, xo x=3 and x=5. We have all 3 solutions, so therefore, x=3, 4, and 5. If you graph those two solutions and using the calculus method, we only see that there are two intersection points at x=4 and x=5 since x=3 is out of the domain. Therefore, the domain for those two functions has to be x≥4.
I set y = x - 4 and moved the entire equation to one side to get:
y^(y + 9) - y^(y + 7) = 0
y^(y + 7) (y^2 - 1) = 0
Therefore:
y^(y+7) = 0 or y^2 = 1
This gives me y = -1, 0 or +1.
Going back to x by adding 4 gives:
x = (3, 4, 5)
x = 5 or 4 or 3
I got x=3,4,5 as solutions.
1^n=1 so if x-4=1, then x=5, easy peasy lemon squeezy
(X-4)^x+3[ (x-4)^2-1]= 0 gives x= 4; 5; 3 solns.
Bro... this equation don't exists, (x-4)^x+3 = (x-4)^x+5 --> x+3 = x+5 --> 3=5 ??????
It does!
You messed up the x=4 case
Why?
@@SyberMaththe base was taken -1,0,1 where it is other than 0.
@@SyberMath 0^(4+3) = 0^(4+5) but you wrote 0^3 = 0^5. Doesn't matter btw, but a bit inaccurate
problem
(x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺⁵⁾
(x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺ ³ ⁺ ²⁾
(x-4)⁽ˣ⁺³⁾ = (x-4)⁽ˣ⁺ ³⁾(x-4)²
Subtract.
(x-4)⁽ˣ⁺³⁾ - (x-4)⁽ˣ⁺ ³⁾(x-4)² = 0
Factor.
(x-4)⁽ˣ⁺ ³⁾ [ 1 - (x-4)² ] = 0
By zero factor property,
(x-4)⁽ˣ⁺ ³⁾ = 0
x = 4
1 = (x-4)²
= x²- 8x + 16
x²- 8x + 15 = 0
x = 5, 3
answer
x ∈ { 3, 4, 5 }
Thanks for taking the time to use superscripts.
you could've really just did:
(x-4)²=1 x-4=±1 x=±1+4 x=3 or x=5
Nice!