Note: This is the Solution of exercise 2.4 from newly published book by PCTB (Punjab Curriculum and Textbook Board, Pakistan) for new 9th session 2025 Onward.
Exercise 2.4
Question # 01: Without using calculator, evaluate the following:
\((i)\) \(\quad\) \(log_2\ 18 – log_2\ 9 \)
\((ii)\) \(\quad\) \(log_2\ 64 + log_2\ 2 \)
\((iii)\) \(\quad\) \(\frac{1}{3} log_3\ 8 -log_3\ 18\)
\((iv)\) \(\quad\) \(2 log\ 2+log\ 25 \)
\((v)\) \(\quad\)\( \frac{1}{3} log_4\ 64+2log_5\ 25\)
\((vi)\) \( \quad\) \(log_3\ 12+log_3\ 0.25 \)
Question 2: Write the following as a single logarithm:
\((i)\) \(\quad\) \(\frac{1}{2} log\ 25+2log\ 3 \)
\((ii)\) \(\quad\) \(log\ 9 -log\ \frac{1}{3}\)
\((iii)\) \(\quad\) \(log_5\ b^2.log_a\ 5^3 \)
\((iv)\) \(\quad \)\(2log_3\ x+log_3\ y\)
\((v)\) \(\quad\) \(4log_5\ x-log_5\ y+log_5\ z \)
\((vi)\) \(\quad\) \(2ln\ a +3ln\ b-4ln\ c \)
Question 3: Expand the following using laws of logarithms:
\((i)\) \(\quad\) \(log\left(\frac{11}{5}\right) \)
\((ii)\) \(\quad\) \(log_{5}\sqrt{8a^{6}}\)
\((iii)\) \(\quad\) \(ln\left(\frac{a^{2} b}{c}\right) \)
\((iv)\) \(\quad \)\(log\left(\frac{xy}{z}\right)^{\frac{1}{9}}\)
\((v)\) \(\quad \) \(ln\sqrt[3]{16x^{3}} \)
\((vi)\) \(\quad \) \(log_{2}\left(\frac{1-a}{b}\right)^{5} \)
Question 4: Find the value of \(x\) in the following equations:
\((i)\) \(\quad\) \(log\ 2+log\ x=1 \)
\((ii)\) \(\quad\) \(log_{2} \ x+log_{2} \ 8=5\)
\((iii)\) \(\quad\) \(( 81)^{x} =( 243)^{x+2} \)
\((iv)\) \(\quad \)\(\left(\frac{1}{27}\right)^{x-6} =27\)
\((v)\) \(\quad\) \(log( 5x-10) =2 \)
\((vi)\) \(\quad\) \(log_{2} \ ( x+1) -log_{2} \ ( x-4) =2 \)
Question 5: Find the values of the following with the help of logarithm table:
\((i)\) \(\quad\) \(\large{\frac{3.68\times 4.21}{5.234} }\)
\((ii)\) \(\quad\) \(4.67\times 2.11\times 2.397\)
\((iii)\) \(\quad\) \(\large{\frac{( 20.46)^{2} \times ( 2.4122)}{754.3} }\)
\((iv)\) \(\quad \)\(\large{\frac{\sqrt[3]{9.364} \times 21.64}{3.21}}\)