Finding Break Even Point

[Everydaylearner]

Hey guys I hope this is in the right spot. I'm working on a Break Even Point equation for finance. I found this similar problem....

(fv÷pv)=((1+r)^4)
=> ($25000 / $11000) = ((1 + r) ^ 4)
=> 2.27272727 = ((1 + r) ^ 4)
=> 1 + r = 1.22782601
=> r = 0.2278601
=> r = 0.2279 = 22.79%

My question is about the highlighted area above. What process/formula do I need to use to get 2.27272727 to equal 1.22782601? Thank you in advance!

 

[Deleted member 4993]

Hey guys I hope this is in the right spot. I'm working on a Break Even Point equation for finance. I found this similar problem....

(fv÷pv)=((1+r)^4)
=> ($25000 / $11000) = ((1 + r) ^ 4)
=> 2.27272727 = ((1 + r) ^ 4)
=> 1 + r = 1.22782601
=> r = 0.2278601
=> r = 0.2279 = 22.79%

My question is about the highlighted area above. What process/formula do I need to use to get 2.27272727 to equal 1.22782601? Thank you in advance!

What would you have done if the equation was:

(1+x)^2 = 10000 ,....... and you needed to solve for 'x'

 

[JeffM]

You are not really asking about a formula. You are asking about approximation techniques.

[imath]\dfrac{25000}{11000}[/imath] does not have an exact decimal solution.

Nor does [imath]r = \sqrt[4]{\dfrac{25,000}{11,000}} - 1.[/imath]

Here is what I suggest you do.

Multiply out 1.227826015 by 1.227826015 by hand. What do you get?

Now multiply that product by itself. What do you get? (There will be a lot of decimal places, 36 to be exact; it may take a while.)

If you round that result to 8 decimal places, does the answer turn out to be close to 2.27272727?

If so, 1.227826015 is actually a pretty good approximation.

And if you subtract 1, then what do you get?

Now actually, you do not need to know how to do approximations because good calculators will do them for you, but if you do not trust calculators, study calculus and numerical methods.