Question

When required the following logarithms may be used.
$\log 2=.3010300,\log 3=.4771213$,
$\log 7=.8450980,\log 11=1.0413927$.
Find the amount of $100$ pounds in $50$ years, at $5$ per cent. compound interest; given $\log 114.674=2.0594650$.

Answer 1

$P=100pounds$

$T=50yrs$

$R=5\%$

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

$A=100(1+\frac{50}{100}{)}^{50}$

$\frac{A}{100}=(\frac{3}{2}{)}^{50}$

$\log \left(\frac{A}{100}\right)=\log \left(\right(\frac{3}{50}{)}^{50}$

$\log \left(A\right)-\log \left(100\right)=50\log \left(\frac{3}{2}\right)$

$\log \left(A\right)-\log \left({10}^2\right)=50(\log 3-\log 2)$

$\log \left(A\right)-2\log \left(10\right)=50(\log 3-\log 2)$

$But,\log 10=1$

$\therefore \log \left(A\right)-2=50(0.4771213-0.3010300)$

$=>\log \left(A\right)=10.804565$

$ThereforeA=antilog\left(10.804565\right)$