>writes his problem not the way he wants >gets angry at people giving solutions to the stated problem
i suspect that you are the low IQ Black person in this thread
69^0 = 1^420
Or if you don't consider 0 a natural number, any of the other solutions in this thread.
1, 1, 1, 1 is also a valid solution because you never specified that they have to be different numbers.
All numbers a and c, as long as their prime factorization contains the same numbers. In short, assuming that a < c, any number a such that a^(b/d) = c, this can be easily extended to negative numbers, imaginary numbers and quaternions upon accounting for -1^2 = 1, -i^2 = -1 etc.
Why not other numbers? Assume that the number a contains at least one prime number that is not in c. Both a^b and c^d have a single unique prime factorization such that upon eliminating all common prime factors (dividing both a^b and c^d by all the prime factors they share), the two numbers are relatively prime to each other. We get e,f where e < f assuming a < c. Ergo, a^b = c^d -> e^b = f^d or e^b/f^d = 1 -> e and f are in fact not relatively prime as we assumed.
0 isn't always treated as a natural number and some texts will refer to N union {0}. It's easier to just say 0 isn't a natural number than it is to say for all natural numbers > 0 all the time in theorems. As a convention, it's perfectly normal to exclude 0.
2 years ago
Anonymous
as a convention, it's not normal to include 0 without stating the inclusion
>8^145 = 32^87
If one is a directly divisor of the other, then that's trivial.
Your example is just two ways of writing 2^(3 * 5 * 29).
It also equals =32768^29 or =536870912^15 >8^145 = 32768^29 >8^145 = 536870912^15 >8^145 = 44601490397061246283071436545296723011960832^3
Come to think of it, that's probably the cheapest way to find more examples.
Choose a base B and an exponent E, do the prime factorization of E and pic from it.
yeah so for even y, x^y = (x^(y/2))^2
it doesn't have to just be 2, either, but y must be divisible by it.
x^y = (x^(y/z))^z
this is real simple but covers most of what people have mentioned
a=c, b=d
'Four' natural numbers. Separate numbers, you low IQ Black folk.
so delete your failure of an op and try writing what you mean you absolute homosexual
>Seething Black person
have a nice day you worthless shitskin.
1,1,1,1 and frick off
>writes his problem not the way he wants
>gets angry at people giving solutions to the stated problem
i suspect that you are the low IQ Black person in this thread
Kek what are you shitskins doing on IQfy? No wonder the question went over your low IQ primitive brains.
you said four natural numbers and did not specify a not c and b not d. go to hell OP
i am not the op but you people are frickin idiots. different letters imply that.
fricking imbecile. then if it would be
a^b=a^b
morons
no they don't, is this some common core shit you learned in zoom school?
x=y
Now what homosexual
>'Four' natural numbers. Separate numbers, you low IQ Black folk.
Not the same thing, ESL.
Post hand.
1,1,1,1
4^3 = 8^2
This exponent offset by 1 generalizes to
[math] ((n^m)*n)^m = (n^m)^{m+1} [/math]
So
", ".join([f"{(n**m)*n}^{m} = {n**m}^{m+1}" for n in range(1,12) for m in range(1,12)])
1^1 = 1^2, 1^2 = 1^3, 1^3 = 1^4, 1^4 = 1^5, 1^5 = 1^6, 1^6 = 1^7, 1^7 = 1^8, 1^8 = 1^9, 1^9 = 1^10, 1^10 = 1^11, 1^11 = 1^12, 4^1 = 2^2, 8^2 = 4^3, 16^3 = 8^4, 32^4 = 16^5, 64^5 = 32^6, 128^6 = 64^7, 256^7 = 128^8, 512^8 = 256^9, 1024^9 = 512^10, 2048^10 = 1024^11, 4096^11 = 2048^12, 9^1 = 3^2, 27^2 = 9^3, 81^3 = 27^4, 243^4 = 81^5, 729^5 = 243^6, 2187^6 = 729^7, 6561^7 = 2187^8, 19683^8 = 6561^9, 59049^9 = 19683^10, 177147^10 = 59049^11, 531441^11 = 177147^12, 16^1 = 4^2, 64^2 = 16^3, 256^3 = 64^4, 1024^4 = 256^5, 4096^5 = 1024^6, 16384^6 = 4096^7, 65536^7 = 16384^8, 262144^8 = 65536^9, 1048576^9 = 262144^10, 4194304^10 = 1048576^11, 16777216^11 = 4194304^12, 25^1 = 5^2, 125^2 = 25^3, 625^3 = 125^4, 3125^4 = 625^5, 15625^5 = 3125^6, 78125^6 = 15625^7, 390625^7 = 78125^8, 1953125^8 = 390625^9, 9765625^9 = 1953125^10, 48828125^10 = 9765625^11, 244140625^11 = 48828125^12, 36^1 = 6^2, 216^2 = 36^3, 1296^3 = 216^4, 7776^4 = 1296^5, 46656^5 = 7776^6, 279936^6 = 46656^7, 1679616^7 = 279936^8, 10077696^8 = 1679616^9, 60466176^9 = 10077696^10, 362797056^10 = 60466176^11, 2176782336^11 = 362797056^12, 49^1 = 7^2, 343^2 = 49^3, 2401^3 = 343^4, 16807^4 = 2401^5, 117649^5 = 16807^6, 823543^6 = 117649^7, 5764801^7 = 823543^8, 40353607^8 = 5764801^9, 282475249^9 = 40353607^10, 1977326743^10 = 282475249^11, 13841287201^11 = 1977326743^12, 64^1 = 8^2, 512^2 = 64^3, 4096^3 = 512^4, 32768^4 = 4096^5, 262144^5 = 32768^6, 2097152^6 = 262144^7, 16777216^7 = 2097152^8, 134217728^8 = 16777216^9, 1073741824^9 = 134217728^10, 8589934592^10 = 1073741824^11, 68719476736^11 = 8589934592^12, 81^1 = 9^2, 729^2 = 81^3, 6561^3 = 729^4, 59049^4 = 6561^5, 531441^5 = 59049^6, 4782969^6 = 531441^7, 43046721^7 = 4782969^8, 387420489^8 = 43046721^9, ...
4^6 = 8^4
They should equal as long as you multiply both exponents by the same number.
Ok, I am moronic. Here's one that matches the criteria: 4^12 = 8^8
lol
4^24 = 8^16
a=x b=yz c=x^y d=z
a=x^y b=zw c=x^z d=yw
a = 16
b = 1
c = 2
d = 4
64^2 = 16^3
9^2 = 3^4
1, 1028304809, 1, 284093859719
Assume n is a perfect square.
Then n^k = (n^(1/2))^k+1
I suspect there's other cases you can solve based on prime roots in rings but I can't do it in my head right now.
Oops, I think you have to treat odd and even k differently.
b=0
c=1
69^0 = 1^420
Or if you don't consider 0 a natural number, any of the other solutions in this thread.
1, 1, 1, 1 is also a valid solution because you never specified that they have to be different numbers.
a = any number
b = 4
c = any number squared
d = 2
44^4 = 1936^2
2^4 = 4^2
{n, 2, n^2, 1}
{n, m, n^m, 1}
{n, m*x, n^(m), x}
Replace n,m,x with whatever.
All numbers a and c, as long as their prime factorization contains the same numbers. In short, assuming that a < c, any number a such that a^(b/d) = c, this can be easily extended to negative numbers, imaginary numbers and quaternions upon accounting for -1^2 = 1, -i^2 = -1 etc.
Why not other numbers? Assume that the number a contains at least one prime number that is not in c. Both a^b and c^d have a single unique prime factorization such that upon eliminating all common prime factors (dividing both a^b and c^d by all the prime factors they share), the two numbers are relatively prime to each other. We get e,f where e < f assuming a < c. Ergo, a^b = c^d -> e^b = f^d or e^b/f^d = 1 -> e and f are in fact not relatively prime as we assumed.
a = 1
b = 2
c = 3
d = 0
0 is a whole number but not a natural number
>0 is... not a natural number
Wrong.
>whole number
Go back to kindergarten.
0 isn't always treated as a natural number and some texts will refer to N union {0}. It's easier to just say 0 isn't a natural number than it is to say for all natural numbers > 0 all the time in theorems. As a convention, it's perfectly normal to exclude 0.
as a convention, it's not normal to include 0 without stating the inclusion
wrong.
I spent a while working it out on paper and the first set I found was 8^145 = 32^87, but I think there are others.
>8^145 = 32^87
If one is a directly divisor of the other, then that's trivial.
Your example is just two ways of writing 2^(3 * 5 * 29).
It also equals =32768^29 or =536870912^15
>8^145 = 32768^29
>8^145 = 536870912^15
>8^145 = 44601490397061246283071436545296723011960832^3
Come to think of it, that's probably the cheapest way to find more examples.
Choose a base B and an exponent E, do the prime factorization of E and pic from it.
For example,
B=5
E=2*7*11
5^(2*7*11) = (5^2)^(7*11) = (5^7)^(2*11) = (5^11)^(2*7) = (5^(7*11))^2 = (5^(2*11))^7 = (5^(2*7))^11
so
5^154 = 25^77 = 78125^22 = 48828125^14 = (5^77)^2 = (5^22)^7 = 6103515625^11
where 5^77 and 5^22 are a bit too long to write in full
a^2 = (a^2)^1
Now plug in any number for a (>2 if you want)
I mean a^b = c^d is
a=b=c=d = false by default ?
2^8 = 4^4
2^16 = 4^8 if you want them all different
3^10 = 9^5. i assume you can see the patterns
3^4 = 9^2
Wow that took 2 seconds.
yeah so for even y, x^y = (x^(y/2))^2
it doesn't have to just be 2, either, but y must be divisible by it.
x^y = (x^(y/z))^z
this is real simple but covers most of what people have mentioned
>(x^(y/z))
being your 'second number'. For instance
5^9 = (5^(9/3))^3
so 5^9 = 125^3
same here. Tho, y wasn't divisible by it, 8^(16/24) is a whole number, so it works.
8^16
(8^(16/24))^24
u^G=A^Y
such that u, G, A, Y are natural numbers satifying the equation