Rewrite the given equation as a^5=2^5 and observe that a represents the five fifth roots of 2^5, each having a magnitude of 2. There are several ways to solve for the five roots, and the easiest and possibly least obvious way is to realize that the five roots lie on a circle of radius 2 centered on the origin of the complex plane. The most obvious root, a=2, lies on the real axis. The five roots are equally spaced on the circle at 2π/5 radian or 72° increments and expressed compactly as a_j=2e^{i2jπ/5}=2[cos(2jπ/5)+i*sin(2jπ/5)],j=0,1,2,3,4.
What a waiste of paper and time . The obvious answer within 3 seconds for any kid is that a =2. Obviously I did not go to Harvard University . I am 76 years old .Too late for that anyhow. Though I do agree that your method is also correct when you involve "i" . Next time you should try to involve Euler with his " e" . It reminds me of mathematical artificial intelligence = AI.
Equation of fifth order has 5 solutions. While real solution 2 is one of them, there are also complex solutions. Saying 2 is solution is incomplete answer.
I agree that my answer "2 "is not complete..There are indeed 4 other correct answers but using "i" is rather artificial by involving the root of 5 . The " ln' approach is a little bit faster. I did not intend to downgrade Harvard University with my reaction. I am way too old and not rich enough to go to Harvard University at age 75.
@christiaanvanhyfte1873 harward university has nothing to do with correctnes and completness of answer. While there are other ways to find all roots of this equation, that does not change the fact that complex solutions exist. Looks like you have some issue with complex numbers. There is nothing artificial with complex numbers, they are quite usefull, for example in solving AC circuits.
Rewrite the given equation as a^5=2^5 and observe that a represents the five fifth roots of 2^5, each having a magnitude of 2. There are several ways to solve for the five roots, and the easiest and possibly least obvious way is to realize that the five roots lie on a circle of radius 2 centered on the origin of the complex plane. The most obvious root, a=2, lies on the real axis. The five roots are equally spaced on the circle at 2π/5 radian or 72° increments and expressed compactly as a_j=2e^{i2jπ/5}=2[cos(2jπ/5)+i*sin(2jπ/5)],j=0,1,2,3,4.
Yep. Seconds instead of 17 minutes. This is the easy way I would solve it.
a^5=32
ln(a^5)=ln(32)
5*ln(a)=ln(32)
a=exp(ln(32)/5
a=2
Thanks
You're welcome! Glad you found it helpful! 👍💖💯
A=2
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Why?
What a waiste of paper and time . The obvious answer within 3 seconds for any kid is that a =2. Obviously I did not go to Harvard University . I am 76 years old .Too late for that anyhow. Though I do agree that your method is also correct when you involve "i" . Next time you should try to involve Euler with his " e" . It reminds me of mathematical artificial intelligence = AI.
Thanks for sharing your alternative perspective 🙏💕😎💯🥰✅💪
Equation of fifth order has 5 solutions. While real solution 2 is one of them, there are also complex solutions. Saying 2 is solution is incomplete answer.
I agree that my answer "2 "is not complete..There are indeed 4 other correct answers but using "i" is rather artificial by involving the root of 5 . The " ln' approach is a little bit faster. I did not intend to downgrade Harvard University with my reaction. I am way too old and not rich enough to go to Harvard University at age 75.
@christiaanvanhyfte1873 harward university has nothing to do with correctnes and completness of answer. While there are other ways to find all roots of this equation, that does not change the fact that complex solutions exist. Looks like you have some issue with complex numbers. There is nothing artificial with complex numbers, they are quite usefull, for example in solving AC circuits.
Thanks