Why dividing a number with a float powered by a big number ends up being a really big number instead of a small number?

Question

I had a problem where I needed to find the result of the division of 2 positive large numbers.

$n1$=$29865^2$ ------- $n2=$$1.5*10^{11}$

When I divided them I got the result of $5.946*10^{19}$

but the correct was $5.946*10^{-3}$ which can be obtained by the division of $1.5*10^{-11}$

Do I always need to convert the exponent to be negative when dividing it by another number? can someone explain me why it happened?

Thank you

@MauroALLEGRANZA I am using the scientific Casio fx-991ES PLUS calculator — Ryan, Mar 21 '22 at 12:37
@MauroALLEGRANZA I don't think that's the case because when I change the exponent to be negative I get the right answer.
so I think it belongs to something mathematically here — Ryan, Mar 21 '22 at 12:40
Most calculators read something like a/b*c as $\frac ab\times c$. — lulu, Mar 21 '22 at 12:42
@MauroALLEGRANZA That should be the answer, not a comment. :) — CiaPan, Mar 21 '22 at 12:43
Do not enter $1.5∗10^{11}$ as a multiplication, but learn to enter it using the built in function. — Jaap Scherphuis, Mar 21 '22 at 12:48
@MauroALLEGRANZA , I got the correct answer using parentheses! thank you, I wonder why it doesn't give the same answer without them, because the multiplication should still make that look as it is still the same variable. but your answer helped me to get the right answer so thank you for that — Ryan, Mar 21 '22 at 13:08
The calculator should have a built-in way to enter numbers in scientific notation, so you enter $1.5*10^{11}$ by typing 1.5E11 where E is the button that is marked with "$\times 10^x$" or sometimes "Exp". — Jaap Scherphuis, Mar 21 '22 at 13:26

score 1 · Answer 1 · answered Mar 21 '22 at 12:53

Try some other ways, equivalent to what you tried 'till now.

For example, compute a reciprocal of $n_2$ and multiply it by $n_1$, like $\frac 1{n_2}\times n_1.$

Or divide $n_2$ by $n_1$ and then take a reciprocal: $\frac 1{n_2/n_1}.$

Or let the whole calculation through logarithm and then exponentiation: $$n_1 : n_2 = 10^{\log n_1 - \log n_2} \\ = 10^{2\log 29865 - (11 + \log 1.5)}.$$

And so on.

If any of those ways gives an answer you expect, you'll know almost for sure the calculator is OK and it's your way of making calculation what is wrong.
Or, if several ways give consistent but unexpected result, then you'll know the calculation is beyond the reach of your device.

Why dividing a number with a float powered by a big number ends up being a really big number instead of a small number?

1 Answers1