[Daily puzzle] log of a sum and sum of logs \ stacker news ~science

[Daily puzzle] log of a sum and sum of logs

671 sats \ 8 comments \ @south_korea_ln 16 Oct science

Can you find $a$, $b$ and $c$ so that this relationship holds?

Previous iteration: #724752

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100 sats \ 5 replies \ @0xbitcoiner 16 Oct

trial and error: a=1 b=2 c=3 I don't remember the log rules anymore :)

100 sats \ 1 reply \ @SimpleStacker 16 Oct

Or any combination of 1, 2, and 3.

You get it because:

Thus, the equation is the same as:

, with

0 sats \ 0 replies \ @0xbitcoiner 16 Oct

👍

0 sats \ 2 replies \ @south_korea_ln OP 16 Oct

That's some lucky trial and error.

Bonus question: can one prove this would be the only solution for

100 sats \ 1 reply \ @SimpleStacker 16 Oct

Haven't worked it out fully, but I think it would go something like this:

Suppose

Then for any

We'd only have to check

because you asked for natural number solutions, and we know 1,2,3 is the smallest such solution.

0 sats \ 0 replies \ @south_korea_ln OP 17 Oct

Only figured it out for symmetric triplets...

Assume

for some

and

. The sum and product of the triplet are:

Sum:

Product:

We set the sum equal to the product:

Assuming

, divide both sides by

This leads to the equation:

Thus,

. For

to be a natural number,

must be a perfect square.

For : The triplet is (x - d, x, x + d) = (1, 2, 3).

No other values of

yield a perfect square for

, because

is not a perfect square for

0 sats \ 1 reply \ @anon 16 Oct

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

__@_'-'

0 sats \ 0 replies \ @anon 16 Oct

Interesting cases:

a = b = c = root(3)

a = b = (1 + root(5))/2, c = 2

__@_'-'