Using EAR formula solve for annual rate when EAR is known

[OB1M]

If the EAR (i) is known how can the equation be rearranged to extract the Annual Rate (r) which is unknown?

 

[MarkFL]

Hello, and welcome to FMH!

I would begin with:

[MATH]i=\left(1+\frac{r}{m}\right)^m-1[/MATH]
Add through by 1:

[MATH]i+1=\left(1+\frac{r}{m}\right)^m[/MATH]
Take the \(m\)th root of both sides:

[MATH](i+1)^{\frac{1}{m}}=1+\frac{r}{m}[/MATH]
Subtract through by 1:

[MATH](i+1)^{\frac{1}{m}}-1=\frac{r}{m}[/MATH]
Multiply through by \(m\) and arrange as:

[MATH]r=m\left((i+1)^{\frac{1}{m}}-1\right)[/MATH]
Using [MATH]i=\frac{I}{100}[/MATH] we may write:

[MATH]r=m\left(\left(\frac{I}{100}+1\right)^{\frac{1}{m}}-1\right)[/MATH]

 

[OB1M]

Hi Mark, thanks for that. Love the graphics, never seen it done like this on a web page. I'm knocking on a bit at 70 and whilst I was taught math to calculus level at college I was poor at it because I needed a practical use for what I was learning and never found one. Now I find myself enjoying financial math purely out of interest. I had already transposed the equation but lacking confidence posted here and you confirmed that my answer was correct.

Question: why can't EAR [MATH]i[/MATH] be used as the basis for interest on all loans that are compounded?

What happens when the exponent numerator and denominator become the same [MATH] n/n[/MATH].

Does the EAR equation break down?

 

[MarkFL]

why can't EAR [MATH]i[/MATH] be used as the basis for interest on all loans that are compounded?

I would assume it can, except perhaps for interest that is continuously compounded.

What happens when the exponent numerator and denominator become the same [MATH] n/n[/MATH].

Does the EAR equation break down?

Where do you see that that can happen?