I cannot figure this out. I’ve tried the calculations on a Javascript calculator online and it’s giving me zero. The answer I want != zero. It’s probably something like .00000000000000006746 or something, but it’s not zero.
Anyway, please divide 14,000,000 by 1.135288e21
and when you get that figure, multiply it by 1.869973e13
Thanks!
*NOTE: This is not homework. I am an adult man, with a job, who is trying to figure out something strictly for my own knowledge.
Wolfram Alpha: 14,000,000 by 1.135288e21 = 1.233167266808069846593992009076111083707394070931781186… × 10^-14
0.23059857939130863710353672372121
I get 0.23059894934
I’m not going to do it for you, but the easiest way is to convert 40,000,000 to the form ae^x. Then it’s just a matter of keeping track of the exponents, where (ae^x)(be^y) = abe^(x+y)
For division, you subtract the 2 exponents.
Note:
ae^x = 40,000,00
ln(ae^x) = 40,000,000
ln(a) +xlne= 40,000,000
Pick any “a” value and you have determined x. It will be easiest if you pick a> 1, but don’t pick a = 1. 
Oh, wait, you can pick a = 1!
Note, though, that two of your input numbers have 7 significant digits, but one of them has only 2 significant digits. Your final result, therefore, can have only 2 significant digits.
I didn’t do the arithmetic myself, but I notice that two posters so far agree on 0.230598 although their digits differ beyond that. Assuming that is correct, then the best you can say is the result is about 0.23 (ETA: Be glad they at least agree on where the decimal point is! 
)
I got this, basically, to more decimal places. I got 0.230598949341489 on my iPhone.
Okay, I did the math, using a HP 32S RPN Scientific hand-held calculator (photo). Yay, RPN!
I got 0.230598949, which agrees with the result beowulff posted above. ETA: And cochrane.
Remember: Only 2 digits count!
OK, now that you have an answer, can we have one?
What are you doing with such huge numbers?
Since you start with something of the order of 10[sup]7[/sup] / 10[sup]21[/sup] and then multiply the answer by 10[sup]13[/sup], you’d expect the answer to be in the vicinity of 10[sup]-1[/sup], or 0.1 - not .00000000000000006746. But the intermediate result was likely flooring your poor calculator…
I have 0.23059894934148867952449070191881 from Windows calculator, but like the boys say, 0.23 is all the accuracy your data supports.
(I’m not trying to be sarcastic with the “Let Me Google.” I just find both the Google Calculater and the lmgtfy site to be nifty!)
Oops. Ignore me, I missed a line…
First work out the division and multiplication first then do the exponents.
First divide 14,000,000 by 1.135288 then multiply by 1.869973.
= 23059894
Then look at the exponent. dividing by 1e21 moves the decimal left 21 spaces, multiplying by 1e13 moves it right 13 spaces.
-21+13=-8 spaces.
23059894 x 1e-8 = .23059894
No, all those zeros before the decimal may count. If it was 1.4e7, then it would have only 2 significant digits for certain.
No need to tax your calculator with all those digits. What you have is (1.4 *1.869973/1.135288)*10^(7+13-21) which gives you the aforementioned 0.23 with the appropriate significant figures.
Wait, wait. How do we know that the 14 000 000 has only two significant figures? It could be 14 million exactly.
ETA: I got 0.230598949 on a TI-30XA.
Never do a problem like this by putting such very large or very small numbers into a calculator, even if it’s a scientific calculator. Calculators are not designed to handle such calculations well. What you should always do is break the values into the coefficients and the exponents. Do the arithmetic first on the coefficients and then on the exponents, and only then should you combine the two numbers. (The coefficient is the number that’s multiplied by a power of ten. The exponent tell which power of ten it is being multiplied by.)
You’re starting with this calculation:
[(1.4 times 10^7) divided by (1.135288 times 10^21)] times (1.869973 times 10^13)
You have the equivalent of (A divided by B) times C. You can rewrite that as (A times C) divided by B. So that calculation is the equivalent of this:
[(1.4 times 1.869973) times (10^7 times 10^13)] divided by (1.135288 times 10^21)
This equals this:
(1.4 times 1.869973) times (10^20) divided by (1.135288 times 10^21))
which equals this:
(2.61379622 times 10^20) divided by (1.135288 times 10^21)
You now have the equivalent of (A times B) divided by (C times D). You can rewrite that as (A divided by C) times (B divided by D). So this is now:
(2.61379622 divided by 1.135288) times (10^20 divided by 10^21)
which equals this:
(2.61379622 divided by 1.135288) times (10^-1)
Now you’re going to have to deal in approximations. 2.61379622 divided by 1.135288 equals approximately 2.305989493. So this is approximately this:
2.305989493 times (10^-1)
And that equals approximately this:
.2305989493
Maybe some calculators don’t handle it well? I would expect a scientific calculator to handle such calculations just fine. Mine does.
Generally, not. This is an ambiguity in this notation.
How many, if any, of the zeroes in 14 000 000 are significant? All of them? None of them? Some of them? How would you know?
That’s why such a number should always be written as 1.4e7 or 1.4 x 10[sup]7[/sup] – but if all of the zeroes are significant, write 1.4000000e7 and if, say, two of the zeroes are significant, write 1.400e7
Thus, one of the specific features of standard scientific notation is to clarify exactly how many of the digits (trailing zeroes in particular) are to be considered significant. In a number like 14 000 000 I think the general consensus would be that the trailing zeroes are not significant. If the originator of the number meant any of them to be significant, he woulda/coulda/shoulda written it in standard scientific form to show that.
Scientists learn to get wary of calculators when mixing extremely large numbers and extremely small numbers. There are many situations where even the best scientific calculator (or even math packages like Mathematica) will give inaccurate results. For example, subtracting two very large numbers that are very close to each other.
The OP’s problem isn’t too bad. It’s just multiplications and divisions, and floating point numbers are suitable for this type of calculation. Still, this problem doesn’t really require a scientific calculator; as Wendell Wagner and others have explained, it’s trivial to break the problem into one that a regular calculator can calculate.
