Question

When required the following logarithms may be used.
$\log 2=.3010300,\log 3=.4771213$,
$\log 7=.8450980,\log 11=1.0413927$.
In how many years will $1000$ pounds become $2500$ pounds at $10$ per cent. compound interest?

Answer 1

$A=P{(1+\frac{R}{100})}^n$

$2500=1000(1+0.1{)}^n=1000(1.1{)}^n$

$\Rightarrow \log \frac{5}{2}=n\log 1.1$

$\Rightarrow n=\frac{1-\log 4}{\log 1.1}=9.61$ years.