Show me how i can enter a long decimal point into the calculator and then have the calculator output an equation and I would love to use my calculator.
Although if you really think that's possible...as a superuser in the algebra forum......
Might wait out for another poster.
If the question is, rather, "how do we find the power, x, so that .6^x= 2.210739 x 10^-7" then take the logarithm, base 10, of both sides: x log(.6)= -7+ log(2.210739...). Are you saying that you what you are writing as "2.210739..." has known digits but too many to put into a calculator?
In that case, the best you can do is enter as many digits as you can, as an approximation to the given number, and get an approximation to x. Doing that with "2.210739" gives
x= 30.000000174627603878991529080142. I don't believe there is any way to prove that x is exactly 30- certainly not without knowing what the other digits are.
Thank you for this great information. What I'm trying to do is generate a shorter number for a long decimal point. I don't need this to work in a calculator, I need this to work in my program, on a much larger scale. We are talking millions of decimal points. So, all I have are my results e.g.(2.623475890234758902345790234598), but I need a way to find the power of equation for this number. Like 2.6232345892345780 = x^y
It seems like encryption is the way to go for large scale shrinking though.
Are logarithms still the way to go?