$\begin{align}&\text{Length of Iron beam}=5.4m-540cm\\[5pt] &\text{Width of Iron beam}=0.36m-36cm\\[5pt] &\text{Depth of Iron beam}=0.22m-22cm\\[5pt] &\text{Mass of one cubic centimetre}=7.86g=\frac {7.86}{1000}kg =0.00786kg\\[5pt] &=540 \times 36\times22\times 0.00786\\[5pt] lg&=540 \times 36\times22 \times 0.00786\\[5pt] &=lg540 \times lg36\times lg22 \times lg0.00786\\[5pt] &=2.7324+1.5563+1.3424+\bar{3}.{8954}\\[5pt] &=3.5265\\[5pt] &=antilog 3.5265\\[5pt] &=\text{Mass of iron beam}\underline{ \underline {3361kg}} \end{align}$
	$\begin{align}g&=\frac{4\Pi^2l}{T^2}\\[5pt] g&=\frac{4\times3.142^2\times1.75}{2.7^2}\\[5pt] lgg&=lg\frac{4\times3.142^2\times1.75}{2.7^2}\\[5pt] lgg&=lg4+lg3.142^2+lg1.75-lg2.7^2\\[5pt] &=0.6021+2\times 0.4972+0.2430-2\times 0.4314\\[5pt] &=0.6021+0.9944+0.2430- 0.8628\\[5pt] &=0.9767\\[5pt] g&=antilog 0.9767\\[5pt] &=\underline{ \underline {9.478}} \end{align}$
	$\begin{align}&\\[5pt] &={\Pi}r_1^2 -{\Pi}r_2^2\\[5pt] &={\Pi}\times0.75^2 -{\Pi}\times 0.07^2\\[5pt] &={\Pi}(0.75^2- 0.07^2)\\[5pt] &={\Pi}\times(0.75+0.07)(0.75-0.07)\\[5pt] &=\text{(i)Area of the remaining part}{\Pi}\times0.82\times0.68\\[5pt] &=3.142\times0.82\times0.68\\[5pt] lg&=lg(3.142\times0.82\times0.68)\\[5pt] lg&=lg3.142+lg0.82+lg0.68\\[5pt] &=0.4972+\bar{1}.9138+\bar{1}.8325\\[5pt] &=0.2435\\[5pt] &=antilog 0.2435\\[5pt] &=\text{(ii)Area }\underline{ \underline {1.752m^2}} \end{align}$
	$\begin{align}&PR=x\\[5pt] &x^2=3.75^2-0.94^2\\[5pt] &=(3.75+0.94)(3.75-0.94)\\[5pt] &=4.69\times2.81\\[5pt] &=\sqrt{4.69\times2.81}\\[5pt] lgx&=lg\sqrt{4.69\times2.81}\\[5pt] &=\frac{1}{2}lg(4.69\times2.81)\\[5pt] &=\frac{1}{2}(lg4.69+lg2.81)\\[5pt] &=\frac{1}{2}(0.6712+0.4487)\\[5pt] &=\frac{1}{2}\times1.1199\\[5pt] &=antilog 0.5599\\[5pt] &=\text{Length of PR}\underline{ \underline {3.63m}} \end{align}$

By studying this lesson you will be able to

Use the table of logarithms to simplify expressions involving products and quotients of powers and roots of numbers 0 and1.