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[Daily puzzle] log of a sum and sum of logs
671 sats \ 8 comments \ @south_korea_ln 16 Oct 2024 science
Can you find $a$, $b$ and $c$ so that this relationship holds?
\log(a+b+c)=\log(a)+\log(b)+\log(c)
Previous iteration: #724752
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100 sats \ 5 replies \ @0xbitcoiner 16 Oct 2024
trial and error: a=1 b=2 c=3 I don't remember the log rules anymore :)
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100 sats \ 1 reply \ @SimpleStacker 16 Oct 2024
Or any combination of 1, 2, and 3.
You get it because:
\log(a) + \log(b) + \log(c) = \log(abc)
Thus, the equation is the same as:
a + b + c = abc , with a, b, c > 0
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0 sats \ 0 replies \ @0xbitcoiner 16 Oct 2024
👍
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0 sats \ 2 replies \ @south_korea_ln OP 16 Oct 2024
That's some lucky trial and error.
Bonus question: can one prove this would be the only solution for a,b,c \in \mathbb{N}?
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100 sats \ 1 reply \ @SimpleStacker 16 Oct 2024
Haven't worked it out fully, but I think it would go something like this:
Suppose a+b+c = abc.
Then for any x,y,z > 0:
(a+x)(b+y)(c+z) > (a+x) + (b+y) + (c+z)
We'd only have to check x,y,z>0 because you asked for natural number solutions, and we know 1,2,3 is the smallest such solution.
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0 sats \ 0 replies \ @south_korea_ln OP 17 Oct 2024
Only figured it out for symmetric triplets...
Assume (a, b, c) = (x - d, x, x + d) for some x and d. The sum and product of the triplet are:
  • Sum:
(x - d) + x + (x + d) = 3x.
  • Product:
(x - d) \cdot x \cdot (x + d) = x \cdot (x^2 - d^2).
We set the sum equal to the product:
3x = x(x^2 - d^2).
Assuming x \neq 0, divide both sides by x:
3 = x^2 - d^2.
This leads to the equation:
x^2 = d^2 + 3.
Thus, x = \sqrt{d^2 + 3}. For x to be a natural number, d^2 + 3 must be a perfect square.
  • For d = 1: 
    x^2 = 1^2 + 3 = 4 \quad \Rightarrow \quad x = 2.
    The triplet is (x - d, x, x + d) = (1, 2, 3).
No other values of d yield a perfect square for x^2, because d^2 + 3 is not a perfect square for d > 1.
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0 sats \ 1 reply \ @anon 16 Oct 2024
Using log laws:
log(a) + log(b) + log(c) = log(abc)
Therefore,
log(abc) = log(a + b + c)
abc = a + b + c
Where,
abc, a, b, c > 0
So from here you just rearrange in terms of a:
a = (b + c) / (bc - 1)
...and pick any value for b and c which are greater than 0. Such as:
b = 2, c = 1
a = (2 + 1) / (2*1 - 1) = 3
__@_'-'
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0 sats \ 0 replies \ @anon 16 Oct 2024
Interesting cases:
a = b = c = root(3)
a = b = (1 + root(5))/2, c = 2
__@_'-'
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