Question

When required the following logarithms may be used.
$\log 2=.3010300,\log 3=.4771213$,
$\log 7=.8450980,\log 11=1.0413927$.
What sum of money at $6$ per cent. compound interest will amount to $1000$ pounds in $12$ years? Given
$\log 106=2.0253059,\log 49697=4.6963292$.

Answer 1

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

$1000=P(1+\mathrm{0.0.6}{)}^{12}=P(1.06{)}^{12}$

$\Rightarrow 3=\log P+12\log 1.06$

$\Rightarrow \log P=3-12\times 0.0253059=2.6963292$

.$\Rightarrow P=496.97$