Thread: Finding the power equation?

Hello!

Quick math question again.

How can I find the power equation of a number such as 2.2107392 × 10^-7. Which equals .6^30?

I'm sure many have asked, sorry if i didn't search the forums properly.

Thank you!

-Rich

Originally Posted by rich2600

Hello!

Quick math question again.

How can I find the power equation of a number such as 2.2107392 × 10^-7. Which equals .6^30?

I'm sure many have asked, sorry if i didn't search the forums properly.

Thank you!

-Rich

Have you tried using your calculator?

CB

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.

Originally Posted by rich2600

Hello!

Quick math question again.

How can I find the power equation of a number such as 2.210739 × 10^-7. Which equals .6^30?

I'm sure many have asked, sorry if i didn't search the forums properly.

Thank you!

-Rich

What is the question you're asking?

It's not at all clear.

I did notice that So, what do you want to know?

Originally Posted by rich2600

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.

You don't need to "enter a long decimal point into the calculator" nor do you want an equation as output. Just enter 0.6 into a calculator (such as the calculator that comes with windows) press the " " key and enter -6. Then look at the result.

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.

Originally Posted by HallsofIvy

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?

Nevermind. Found a better alternative. Remainders.

I think most of us who have looked at this thread, especially those who have responded, (we) would like to know what it was that you were after here.

Sorry I was to subtle earlier. Hope this clears up what i was looking for.
Tip: It has arrows pointing at it.