Hello, it's that time again where I post to you lovely people for some help / advice. This time around I have the full working and correct answer but need a part of the working to be explained to me. I'll post the question & the answer and then identify what I need help understanding. As, always, thank you all in advance! :)
Question
An investment account offers an interest rate of 6% per annum, compounded half yearly. Assuming a single deposit of money is placed in this investment account, how long would it take for £29000 to amount to £68000, assuming that £1000 is withdrawn at the end of the first year and thereafter no further withdrawals are made?
Answer

At = A0 (1+r / 100m)^mt
With this in mind: At= 68,000 A0 = 29,000 t = ? R = 6% m = 2

First we must calculate the value at A1 and deduct the £1000 withdrawal.
A1 = 29000 * (1 + 6/200) ^ 2
= 29000 * 1.03^2 - 1000
= 29766.1

Now we have the value of A1 after the withdrawal is made we can reinvest that amount for t-1 years to achieve a final amount of £68,000.

68000 = 29776.1 * 1.03 ^2t-1
At this stage we can then divide both sides by 29766.1 to help solve the problem:
68000 / 29766.1 = 1.03^2t-2

log 68000 / 29766.1 = (2t-2) log 1.03
2t-2 = (log 68000/29766.1) / log 1.03 = 28
2t = 30
t = 15

The part of the question that is bold and italics is where I need some help in understanding the solution. Why does 2t-1 become 2t-2 after the division is carried out?

The quantity t represents the number of 6-month periods, not years. Note that the interest rate paid is 3% per period, not 6%, so you know that t is defined to be 6-month intervals.

Hence the first bold formula should be:


68000 = 29776.1\ \times \ 1.03^{(t-2)}


because two 6-month times have passed at that point.

Excellent! I can't believe I haven't picked that up from reading the course material! D'oh!
Thank you very much, appreciate the time taken to help me out! :)