Find the compound interest on Rs. 12000 for 10 years at the rate of 12% per annum compounded annually in Rs.

We know that the amount A at the end of n years at the rate of $r$ per annum when the interest is compounded annually is given by
$A=P{(1+\frac{r}{100})}^n$
Here, $P=Rs.12000$, $r=12$ and $n=10$
$\therefore$ $A=Rs.\left[12000{(1+\frac{12}{100})}^{10}\right]$
$=Rs.\left[12000{(1+\frac{3}{25})}^{10}\right]$
$=Rs.\left[12000{\left(\frac{25+3}{25}\right)}^{10}\right]$
$=Rs.\left[12000{\left(\frac{28}{25}\right)}^{10}\right]$
Now $A=Rs.12000{\left(\frac{28}{25}\right)}^{10}$
$\Rightarrow$ $\log A=\log 12000+10(\log 28-\log 25)$
$=4.0792+10(1.4472-1.3979)$
$=4.0792+0.493=4.5722$
$\Rightarrow$ $A=antilog\left(4.5722\right)=37350$.
So, the amount after 10 years is Rs. 37350.
Hence, Compound interest$=Rs.(37350-12000)=Rs.25350$
Ans: Rs.25350