Simple Arithmetic Puzzle 4. Or is it?

Question

\begin{align}56\times0.7&=367\\0.7\times56&=367\\12\times34&=601\\1.2\times3.4&=141\\\sqrt2\times\sqrt3&=90\\\frac1{\sqrt2}\times\frac1{\sqrt3}&=-90\\1\times1&=\quad?\end{align}

Can you find the value of the question mark?

Answer 1

The rule for the multiplication is

$a \times b \equiv [ 100 \ln(ab) ] = [ 100 (\ln a + \ln b) ]$
where the brackets $[ ]$ denote the Nearest Integer Function.

Examples

$56 \times 0.7 = [ 100 \ln(39.2) ] = [ 366.86\ldots ] = 367$
$12 \times 34 = [100 \ln(408)] = [601.126\ldots] = 601$
$\sqrt{2} \times \sqrt{3} = [100 \ln(\sqrt{6})] = [89.587\ldots] = 90$

How I found this solution

The $5$th and $6$th equations give the hint that this may have something to do with taking logs. Generally, if I see a function where the reciprocal argument yields the negative solution that is what I would try. Also, the number on the right seems to increase as the product of the numbers on the left increases. After messing around with logs it didn't take long to find a multiplicative factor that worked.

Answer 2

I noticed something. Is it worth mentioning? I don't know. But here it is anyway:

If you actually carry out the multiplication, multiply by $10$, and then subtract by/from the answer that has been put up, you will get a number near to a square number (subtract by if larger; subtract from if smaller).

Supporting Examples:

$$\begin{align}56\times 0.7 &= 39.2 \tag*{$\big($also equals $0.7\times 56\big)$}\\ 39.2\times 10 &= 392 \\ 392-367 &= 25 = 5^2\end{align}$$ $$\begin{align}12\times 34 &= 408 \\ 408\times 10 &= 4080 \\ 4080-601 &= 3479=59^2-2\end{align}$$ $$\begin{align}1.2\times 3.4 &= 4.08 \\ 4.08\times 10 &= 40.8 \\ 141-40.8 &= 100.2 = 10^2+0.2\end{align}$$

And then you got the radicals:

$$\begin{align}\sqrt 2\times \sqrt 3&\simeq 2.4494897427831781\; (1) \\ (1)\times 10 &= 24.494897427831781\; (2) \\ 90-(2) &= 65.505102572168219=8^2+1.505102572168219\end{align}$$

Also,

In the puzzle, it appears that if $a=b$ then $1\div a=-b$. Let $a=\sqrt2\times \sqrt3$ and $b=90$.

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So, my best bet from this observation is that:

$$1\times 1 = 9$$

Since

$$\begin{align}1\times 1 &= 1 \\ 1\times 10 &= 10 \\ 10-9 &= 1 = 1^2\end{align}$$ And I will assume that since $1\times 1 = 1$, then in the puzzle, $1\times 1\neq 1$. We also have that $10\times 9 - 9=9^2$ so that is related.