How To Solve A Weird Exponential Equation | Homemade
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- Опубліковано 16 вер 2024
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3^x+5^x=0
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(3/5)^x should be expressed as [(3/5)*(2kπi)]^x where k is an integer. Then we can get a more general solution. Without putting k, the solution is the specific case when k = 1.
I tried this and the resulting x is not a solution to the equation. You get x = i PI /( ln(3/5) + 2 PI i k ) (the numerator comes from RHS where ln(-1) = i PI.
x is a solution for k = 0, but is not a solution for all other values of k.
Should be [(3/5)*exp(2kπi)]^x , I forgot to type exp()
@@82rah I typed wrongly, it should be [(3/5)*exp(2kπi)]^x
1 = exp(2kπi)
@@82rah the solution in the video put (3/5)^x = [(3/5)*exp(2πi)]^x which is the case when k = 1. If the solution is correct only when k = 0, then the suggested solution in this video when k = 1 is not correct too.
x=iπ/ln(5/3)
2:26 - why did you add here e^(2pi * i)? Without adding this I get the answer x=(i*pi)(1+2n)/(ln5-ln3), n is integer.
I agree. x=(i*pi)(1+2n)/(ln5-ln3) is the solution. Adding the e^(2pi * i) on LHS leads to an x that is not a solution. I do not understand why this is.
I agree. That was a mistake
I think you left out a multiple of pi in the imaginary part of the solution.
😮
This is not correct. Calculate x for n=0 get x = .496717 - .0128544 i then 3^x + 5^x = 3.949502 - .0703846 i This arises from the exp (2 PI i ) on the LHS. w/o this the solution is x = i PI ( 2 n + 1 )/(ln(5/3) for n= 0 x =6.140029 i 3^x + 5^x = 0 x is soln for other n: if x = -3 x = =30.750147 i and again 3^x + 5^x = 0
Nonono x is pure imaginary! x=(2n-1) i pi /(ln 5 - ln 3)
Why not just solving a^x + 1 = 0?
Great video anyway!
x=(2n-1) i pi /(ln 5 - ln 3)
Can you depict that imaginary solution in a graph?
not sure
@@SyberMath So, what's the point solving an equation when you can't depict the solutions?
Impossible
x=iπ/ln(5/3)
This is one of the solution when k = 0 and n = 0.