1. ## [SOLVED] Consumer Mathematics

I placed $1200 in an bank investment which paid 8% p.a compound interest. At the same time, I placed$1800 in a bank which paid 4% p.a compound interest. During which year would the $1200 investment be worth more than the$1800 investment. I can do this question by listing the figures at the end of every year, but I wanted to know if there is way to solve this using algebra?

2. ## Compound Interest

Hello requal
Originally Posted by requal
I placed $1200 in an bank investment which paid 8% p.a compound interest. At the same time, I placed$1800 in a bank which paid 4% p.a compound interest. During which year would the $1200 investment be worth more than the$1800 investment. I can do this question by listing the figures at the end of every year, but I wanted to know if there is way to solve this using algebra?
The compound interest formula gives the amount A after n years as:

$A = P(1+r)^n$

So into this formula you have to plug in each of the values of $P$ and $r$, equate the two and solve for $n$ to find the point at which the two amounts become equal.

$1200(1.08)^n = 1800(1.04)^n$

$\Rightarrow 1.08^n = 1.5(1.04)^n$

$\Rightarrow \left(\frac{1.08}{1.04}\right)^n = 1.5$

$\Rightarrow 1.03846^n = 1.5$

Now - and here's the key step - take logs of both sides (it doesn't matter which base of logs you use):

$n\log(1.03846) = \log(1.5)$

$\Rightarrow n = \frac{\log(1.5)}{\log(1.03846)}$

$= 10.74$

So by the end of year 11, the first investment will be worth more than the second.