This question has two related parts — when the amount of interest in the payment is equal to the equity amount in the payment; and when the total amount of interest paid equals the total amount of principle paid. They are around the same time, but not exactly.

Interest payment equals principle payment

To calculate this, we need to determine when the:

interest paid in month p = 0.5 × monthly payment

That is, when:

`(Lr(1-(1+r)^(p-n)))/(1-(1+r)^-n)` ` = 0.5xx(Lr)/(1-(1+r)^-n)`

Simplifying this (by multiplying throughout by `(1-(1+r)^-n)` and dividing by `Lr` gives:

`1-(1+r)^(p-n) = 0.5`

`(1+r)^(p-n) = 0.5`

Using our example from before, with:

L = 270000

r = 0.08/12 = 0.0066667

p = unknown

n = 360

We only need `r` and `n`, so we substitute:

`(1+0.00666667)^(p-360) = 0.5`

Taking log of both sides:

`(p-360)log(1+0.00666667) ` `= log(0.5)`

Solving for `p`:

`p ` `= (log(0.5))/log(1+0.00666667) + 360` `~~256`

So the month where the amount paid in interest and the amount paid in principle is (very close to) the same, is the 256th month.

Total Interest paid equals Total Principle paid

The interest part of the total amount paid so far will equal the principal paid so far when the the balance owing equals the equity (the amount of the house we own).

We saw earlier that:


The amount of equity we have in the house is simply the house value minus what we still owe. So:

`text(Equity)=L - (L[1-(1+r)^(p-n)])/(1-(1+r)^-n)`

Setting these equal gives:

`(L[1-(1+r)^(p-n)])/(1-(1+r)^-n)` `=L - (L[1-(1+r)^(p-n)])/(1-(1+r)^-n)`

We need to solve for `p`, the number of payments already made.

Multiplying throughout by `1-(1+r)^-n` gives:

`L[1-(1+r)^(p-n)]` `=L[1-(1+r)^-n]` ` - L[1-(1+r)^(p-n)]`

Adding `L[1-(1+r)^(p-n)]` to both sides then cancelling the `L`'s gives:

`2[1-(1+r)^(p-n)]` `=[1-(1+r)^-n]`


`2[1 - ((1+r)^p)/(1+r)^n] =1-1/((1+r)^n)`

Multiplying throughout by `(1+r)^n` gives:

`2[(1+r)^n - (1+r)^p]` ` =(1+r)^n - 1`

`(1+r)^n - 2(1+r)^p = - 1`

Tidying this up and getting the `(1+r)^p` term on the left hand side:

`(1+r)^p = 0.5[(1+r)^n+1]`

Taking log of both sides:

`log(1+r)^p = log[0.5((1+r)^n+1)]`

This is equivalent to:

`p log(1+r) = log[0.5((1+r)^n+1)]`


`p = (log[0.5((1+r)^n+1)])/log(1+r)`

In the example given earlier on this page, we had:

L = 270000

r = 0.08/12 = 0.0066667

p = unknown

n = 360

Therefore, the required time is:

`p = (log[0.5((1+0.08/12)^360+1)])/log(1+0.08/12)` ` = 280.959 ~~ 281`

This means the principal equals the equity on the 281st payment, i.e. the 4th payment of the 23rd year.

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