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
trial and error: a=1 b=2 c=3
I don't remember the log rules anymore :)
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
👍
That's some lucky trial and error.
Bonus question: can one prove this would be the only solution for a,b,c∈N?
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.
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:
We set the sum equal to the product:
Assuming x=0, divide both sides by x:
This leads to the equation:
Thus, x=d2+3. For x to be a natural number, d2+3 must be a perfect square.
The triplet is (x - d, x, x + d) = (1, 2, 3).
No other values of d yield a perfect square for x2, because d2+3 is not a perfect square for d>1.
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
__@_'-'
Interesting cases:
a = b = c = root(3)
a = b = (1 + root(5))/2, c = 2
__@_'-'