Question 1: Simplify each of the following:

$\displaystyle \text{(i) } \log x^5 - \log x^4$     $\displaystyle \text{(ii) } \log x^5 \div \log x^4$      $\displaystyle \text{(iii) } \frac{\log 27}{\log 9 }$

$\displaystyle \text{(iv) } \log_6 72 - \log_6 2$     $\displaystyle \text{(v) } \frac{\log 27 - \log 9}{\log 81}$

$\displaystyle \text{(i) } \log x^5 - \log x^4 = \log \frac{x^5}{x^4} = \log x$

$\displaystyle \text{(ii) } \log x^5 \div \log x^4 = \frac{\log x^5}{\log x^4} = \frac{5 \log x}{4 \log x} = \frac{5}{4}$

$\displaystyle \text{(iii) } \frac{\log 27}{\log 9 } = \frac{\log 3^3}{\log 3^2} = \frac{3}{2}$

$\displaystyle \text{(iv) } \log_6 72 - \log_6 2 = \log_6 \frac{72}{2} = \log_6 36 = \log_6 6^2 = 2$

$\displaystyle \text{(v) } \frac{\log 27 - \log 9}{\log 81} = \frac{3 \log 3 - 2 \log 3}{4 \log 3} = \frac{1}{4}$

$\displaystyle \\$

Question 2: Prove the following:

$\displaystyle \text{(i) } \log 108 = 3 \log 3 + 2 \log 2$     $\displaystyle \text{(ii) } \log 75 = \log 3 + 2 \log 5$

$\displaystyle \text{(iii) } \log 72 = 2 \log 3 + 3 \log 2$     $\displaystyle \text{(iv) } 2 \log 5 + \log 8 - \frac{1}{2} \log 4 = 2$

$\displaystyle \text{(v) } \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = 0$     $\displaystyle \text{(vi) } \log 2 + 16 \log \frac{16}{15} + 12 \log \frac{25}{24} +7 \log \frac{81}{80} = 1$

$\displaystyle \text{(vii) } \log 2 + 2 \log 5 - \log 3 - 2 \log 7 = \log \frac{50}{147}$

$\displaystyle \text{(viii) } 2\log \frac{11}{13} + \log \frac{130}{77} - \log \frac{55}{91} = \log 2$

$\displaystyle \text{(ix) } \log \frac{75}{16} + \log \frac{32}{243} - 2 \log \frac{5}{9} = \log 2$

$\displaystyle \text{(x) } 7 \log \frac{10}{9} + 3 \log \frac{81}{80} - 2 \log \frac{25}{24} = \log 2$

$\displaystyle \text{(xi) } \log (1^{1/5} + 32^{1/5} + 243^{1/5}) = \frac{1}{5} (\log 1 + \log 32 + \log 243 )$

$\displaystyle \text{(xii) } 7 \log_2 \frac{16}{15} + 5 \log_2 \frac{25}{24} + 3 \log_2 \frac{81}{80} = 1$

$\displaystyle \text{(xiii) } \log \Big( \frac{x^2}{yz} \Big) + \log \Big( \frac{y^2}{zx} \Big) + \log \Big( \frac{z^2}{xy} \Big) = 0$

$\displaystyle \text{(i) } \log 108 = 3 \log 3 + 2 \log 2$

$\displaystyle \text{LHS } = \log 108 = \log (27 \times 4) = \log 27 + \log 4 = 3 \log 3 + 2 \log 2 = \text{ RHS }$

$\displaystyle \text{(ii) } \log 75 = \log 3 + 2 \log 5$

$\displaystyle \text{LHS } = \log 75 = \log (25 \times 3) = \log 25 + \log 3 = 2 \log 5 + \log 3 = \text{ RHS }$

$\displaystyle \text{(iii) } \log 72 = 2 \log 3 + 3 \log 2$

$\displaystyle \text{LHS } = \log (8 \times 9) = \log 8 + \log 9 = 3 \log 2 + 2 \log 3 = \text{ RHS }$

$\displaystyle \text{(iv) } 2 \log 5 + \log 8 - \frac{1}{2} \log 4 = 2$

$\displaystyle \text{LHS } = \log 25 + \log 8 - \log 2 = \log \frac{25 \times 8}{2} = \log 100 = \log 10^2 = 2 = \text{ RHS }$

$\displaystyle \text{(v) } \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = 0$

$\displaystyle \text{LHS } = \log \frac{4}{7} + \log \frac{33}{18} - \log \frac{22}{21} = \log \Big( \frac{4}{7} \times \frac{33}{18} \times \frac{21}{22} \Big) = \log 1 = 0$

$\displaystyle \text{(vi) } \log 2 + 16 \log \frac{16}{15} + 12 \log \frac{25}{24} +7 \log \frac{81}{80} = 1$

$\displaystyle \text{LHS } = \log 2 + \log ( \frac{16}{15} )^{16} + \log ( \frac{25}{24} )^{12} + \log ( \frac{81}{80} )^7$

$\displaystyle = \log \Big( 2 \times ( \frac{16}{15} )^{16} \times ( \frac{25}{24} )^{12} \times ( \frac{81}{80} )^7 \Big) = \log 10 = 1$

$\displaystyle \text{(vii) } \log 2 + 2 \log 5 - \log 3 - 2 \log 7 = \log \frac{50}{147}$

$\displaystyle \text{LHS } = \log 2 + 2 \log 5 - \log 3 - 2 \log 7 = \log \Big( \frac{2 \times 5^2}{3 \times 7^2} \Big) = \log \frac{50}{147}$

$\displaystyle \text{(viii) } 2\log \frac{11}{13} + \log \frac{130}{77} - \log \frac{55}{91} = \log 2$

$\displaystyle \text{LHS } = 2\log \frac{11}{13} + \log \frac{130}{77} - \log \frac{55}{77} = \log \Big( ( \frac{11}{13} )^2 \times \frac{130}{77} \times \frac{77}{91} \Big) = \log \frac{110}{55} = \log 2$

$\displaystyle \text{(ix) } \log \frac{75}{16} + \log \frac{32}{243} - 2 \log \frac{5}{9} = \log 2$

$\displaystyle \text{LHS } = \log \frac{75}{16} + \log \frac{32}{243} - 2 \log \frac{5}{9} = \log ( \frac{75}{16} \times \frac{32}{243} \times ( \frac{9}{5})^2) = \log 2$

$\displaystyle \text{(x) } 7 \log \frac{10}{9} + 3 \log \frac{81}{80} - 2 \log \frac{25}{24} = \log 2$

$\displaystyle \text{LHS } = 7 \log \frac{10}{9} + 3 \log \frac{81}{80} - 2 \log \frac{25}{24} = \log ( ( \frac{10}{9} )^7 \times ( \frac{81}{80} )^3 \times ( \frac{24}{25} )^2 ) = \log 2$

$\displaystyle \text{(xi) } \log (1^{1/5} + 32^{1/5} + 243^{1/5}) = \frac{1}{5} (\log 1 + \log 32 + \log 243 )$

$\displaystyle \text{LHS } = \log (1^{1/5} + 32^{1/5} + 243^{1/5}) = \log (1 + 2 + 3 ) = \log 6$

$\displaystyle \text{RHS } = \frac{1}{5} (\log 1 + \log 32 + \log 243 ) = \log 1^{1/5} + \log 32^{1/5} + \log 243^{1/5}$

$\displaystyle = \log 1 + \log 2 + \log 3 = \log (1 \times 2 \times 3 ) = \log 6$

Hence LHS = RHS

$\displaystyle \text{(xii) } 7 \log_2 \frac{16}{15} + 5 \log_2 \frac{25}{24} + 3 \log_2 \frac{81}{80} = 1$

$\displaystyle \text{LHS } = 7 \log_2 \frac{16}{15} + 5 \log_2 \frac{25}{24} + 3 \log_2 \frac{81}{80}$

$\displaystyle = \log_2 \Big( ( \frac{16}{15} )^7 \times ( \frac{25}{24} )^5 \times ( \frac{81}{80} )^3 \Big) = \log_2 2 = 1$

(xiii) $\displaystyle \log \Big( \frac{x^2}{yz} \Big) + \log \Big( \frac{y^2}{zx} \Big) + \log \Big( \frac{z^2}{xy} \Big) = 0$

$\displaystyle \text{LHS } = \log \Big( \frac{x^2}{yz} \Big) + \log \Big( \frac{y^2}{zx} \Big) + \log \Big( \frac{z^2}{xy} \Big)$

$\displaystyle = \log \Big( \frac{x^2}{yz} \times \frac{y^2}{zx} \times \frac{z^2}{xy} \Big) = \log \frac{x^2.y^2.z^2}{x^2.y^2.z^2} = \log 1 = 0$

$\displaystyle \\$

Question 3: Express each of the following as the logarithm of a single number:

$\displaystyle \text{(i) } \log 3 + 2$     $\displaystyle \text{(ii) } 2 \log 3 + 3$      $\displaystyle \text{(iii) } 1 + \frac{1}{3} \log 27$

$\displaystyle \text{(iv) } \frac{1}{2} \log 4 + \frac{1}{4} \log 81 + 3 \log 2 - \log 6+2$

$\displaystyle \text{(v) } \frac{1}{2} \log 9 + 2 \log 3 - \log 6 + \log 2 - 2 = \log 3 + \log 9 - \log 6 + \log 2 - \log 100$

$\displaystyle \text{(vi) } 2 + \frac{1}{2} \log_{10} 9 - 2 \log_{10} 5$

$\displaystyle \text{(i) } \log 3 + 2 = \log 3 + \log 100 = \log 300$

$\displaystyle \text{(ii) } 2 \log 3 + 3 = \log 3^2 + \log 1000 = \log 9000$

$\displaystyle \text{(iii) } 1 + \frac{1}{3} \log 27 = \log 10 + \log 27^{1/3} = \log 10 + \log 3 = \log 30$

$\displaystyle \text{(iv) } \frac{1}{2} \log 4 + \frac{1}{4} \log 81 + 3 \log 2 - \log 6+2$

$\displaystyle = \log 2 + \log 3 + \log 8 - \log 6 + \log 100 = \log \frac{2 \times 3 \times 8 \times 100}{6} = \log 800$

$\displaystyle \text{(v) } \frac{1}{2} \log 9 + 2 \log 3 - \log 6 + \log 2 - 2 = \log 3 + \log 9 - \log 6 + \log 2 - \log 100$

$\displaystyle = \log \frac{3 \times 9 \times 2}{6 \times 100} = \log \frac{9}{100}$

$\displaystyle \text{(vi) } 2 + \frac{1}{2} \log_{10} 9 - 2 \log_{10} 5 = \log_{10} 100 + \log_{10} 3 - \log_{10} 25 = \log_{10} \frac{100 \times 3}{25} = \log 12$

$\displaystyle \\$

Question 4: Evaluate the following:

$\displaystyle \text{(i) } 2 \log 5 + \frac{1}{3} \log 64$     $\displaystyle \text{(ii) } \log 21 + \log 4 + 2 \log 5 - \log 3 - \log 7$

$\displaystyle \text{(iii) } \log 15 + 2 \log 0.5 + 3 \log 2 - \log 3 - \log 5$     $\displaystyle \text{(iv) } \log_3 5 \times \log_{25} 27$

$\displaystyle \text{(v) } \log_4 \{\log_{\sqrt{2}} (\log_3 81) \}$     $\displaystyle \text{(vi) } \log_3 \sqrt{\frac{2}{3}} - \log_3 (\log_3 9))$

$\displaystyle \text{(i) } 2 \log 5 + \frac{1}{3} \log 64 = \log 25 + \log 4 = \log 100 = \log 10^2 = 2$

$\displaystyle \text{(ii) } \log 21 + \log 4 + 2 \log 5 - \log 3 - \log 7 = \log \frac{21 \times 4 \times 25}{3 \times 7} = \log 100 = 2$

$\displaystyle \text{(iii) } \log 15 + 2 \log 0.5 + 3 \log 2 - \log 3 - \log 5 = \log \frac{15 \times 0.5 \times 0.5 \times 8}{3 \times 5} = \log 2$

$\displaystyle \text{(iv) } \log_3 5 \times \log_{25} 27 = \log_3 5 \times \log_{5^2} 3^3 = \log_3 5 \times \frac{3}{2} \log_5 3 = \frac{3}{2}$

$\displaystyle \text{(v) } \log_4 \{\log_{\sqrt{2}} (\log_3 81) \} = \log_4 \{ \log_{\sqrt{2}} (\log_3 3^4 ) \}$

$\displaystyle = \log_4 (\log_{\sqrt{2}} 4 ) = \log_4 (\log_{\sqrt{2}} (\sqrt{2})^4 = \log_4 4 = 1$

$\displaystyle \text{(vi) } \log_3 \sqrt{\frac{2}{3}} - \log_3 (\log_3 9)) = \log_3 \sqrt{6} + \log_3 \sqrt{\frac{2}{3}} - \log_3 2$

$\displaystyle = \log_3 \Big( \frac{\sqrt{6} \times \sqrt{2}}{2 \times \sqrt{3}} \Big) = \log_3 1 = 0$

$\displaystyle \\$

Question 5: Prove that:

$\displaystyle \text{(i) } \log_{10} 4 \div \log_{10} 2 = \log_3 9$      $\displaystyle \text{(ii) } \log_{10} 25 + \log_{10} 4 = \log_5 25$

$\displaystyle \text{(i) } \log_{10} 4 \div \log_{10} 2 = \log_3 9$

$\displaystyle \text{LHS } = 2 \log_{10} 2 \div \log_{10} 2 = 2$

$\displaystyle \text{RHS } = \log_3 3^2 = 2 \log_3 3 = 2$. Hence proved.

$\displaystyle \text{(ii) } \log_{10} 25 + \log_{10} 4 = \log_5 25$

$\displaystyle \text{LHS } = 2 \log_{10} 5 + 2 \log_{10} 2 = 2(\log_{10} 5+ \log_{10} 2) = 2 \log_{10} 10 = 2$

$\displaystyle \text{RHS } = \log_5 25 = \log_5 5^2 = 2 \log_5 5 = 2$. Hence proved.

Question 6: $\displaystyle \text{If } x = 100^a, y = 10000^b \text{ and } z = 10^c$, express $\displaystyle \log \Big( \frac{10\sqrt{7}}{x^2z^3} \Big)$ in terms of $\displaystyle a, b \ and \ c$.

$\displaystyle x = 100^a \Rightarrow \log x = 2a$

$\displaystyle y = 10^{4b} \Rightarrow \log y = 4b$

$\displaystyle z = 10^c \Rightarrow \log z = c$

$\displaystyle \log \Big( \frac{10\sqrt{7}}{x^2z^3} \Big)$

$\displaystyle = \log 10 + \log \sqrt{y} - \log x^2 - \log z^3$

$\displaystyle = 1 + \frac{1}{2} \log y - 2 \log x - 3 \log z$

$\displaystyle = 1 + \frac{1}{2} . 4b - 2. 2a - 3 c$

$\displaystyle = 1 +2b-4a-3c$

$\displaystyle \\$

Question 7: $\displaystyle \text{If } 2 \log x \text{ and } 2 \log y -2 = 0 \text{ , prove that } x^2y^3 = 100$

$\displaystyle 2 \log x \text{ and } 2 \log y -2 = 0$

$\displaystyle \Rightarrow \log x^2 + \log y^3 - \log 100 = \log 1$

$\displaystyle \Rightarrow \log \Big( \frac{x^2y^3}{100} \Big) = \log 1$

$\displaystyle \Rightarrow \frac{x^2y^3}{100} = 1$

$\displaystyle \Rightarrow x^2y^3 = 100$

$\displaystyle \\$

Question 8: $\displaystyle \text{If } \log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log a + \log b) \text{ , prove that } a^2 + b^2 = 7ab$

$\displaystyle \log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log a + \log b)$

$\displaystyle \Rightarrow \log \Big( \frac{a+b}{3} \Big) = \frac{1}{2} (\log ab)$

$\displaystyle \Rightarrow \log \Big( \frac{a+b}{3} \Big) = (\log (ab)^{\frac{1}{2}})$

$\displaystyle \Rightarrow \frac{a+b}{3} = (ab)^{\frac{1}{2}}$

$\displaystyle \Rightarrow a^2 + b^2 +2ab = 9 ab$

$\displaystyle \Rightarrow a^2 + b^2 = 7 ab$

$\displaystyle \\$

Question 9: $\displaystyle \text{If } a^{2x-3} b^{2x} = a^{6-x} b^{5x} \text{ , prove that } 3 \log a = x \log \frac{a}{b}$

$\displaystyle a^{2x-3} b^{2x} = a^{6-x} b^{5x}$

$\displaystyle \Rightarrow \frac{a^{2x-3}}{a^{6-x}} = \frac{b^{5x}}{b^{2x}}$

$\displaystyle \Rightarrow a^{2x-3-6+x} = b^{5x-3x}$

$\displaystyle \Rightarrow a^{3x-9} = b^{3x}$

$\displaystyle \Rightarrow (a^{x-3})^3 = (b^x)^3$

$\displaystyle \Rightarrow a^{x-3} = b^x$

$\displaystyle \\$

Question 10: $\displaystyle \text{If } \log \Big( \frac{a+b}{2} \Big) = \frac{1}{2} (\log a + \log b) \text{ , prove that } a = b$.

$\displaystyle \log \Big( \frac{a+b}{2} \Big) = \frac{1}{2} (\log a + \log b)$

$\displaystyle \Rightarrow \log \Big( \frac{a+b}{2} \Big) = (\log a^{\frac{1}{2}} + \log b^{\frac{1}{2}})$

$\displaystyle \Rightarrow \log \Big( \frac{a+b}{2} \Big) = (\log (ab)^{\frac{1}{2}})$

$\displaystyle \Rightarrow \frac{a+b}{3} = (ab)^{\frac{1}{2}}$

$\displaystyle \Rightarrow \frac{a^2 + b^2 + 2ab}{4} = ab$

$\displaystyle \Rightarrow a^2 + b^2 -2ab = 0$

$\displaystyle \Rightarrow (a-b)^2 = 0$

$\displaystyle \Rightarrow a-b = 0 \ or \ a = b$

$\displaystyle \\$

Question 11: $\displaystyle \text{If } \log_{10} a + \frac{1}{2} \log_{10} b = 1 \text{ , prove that } ba^2 = 100$

$\displaystyle \log_{10} a + \frac{1}{2} \log_{10} b = 1$

$\displaystyle \Rightarrow \log_{10} a + \frac{1}{2} \log_{10} b = \log_{10} 10$

$\displaystyle \Rightarrow \log_{10} ab^{1/2} = \log_{10} 10$

$\displaystyle \Rightarrow ab^{\frac{1}{2}} = 10$

$\displaystyle \Rightarrow a^2b = 100$

$\displaystyle \\$

Question 12: $\displaystyle \text{If } \log \frac{a-b}{2} = \frac{1}{2} (\log a + \log b) \text{ , prove that } a^2 + b^2 = 6ab$

$\displaystyle \text{If } \log \frac{a-b}{2} = \frac{1}{2} (\log a + \log b)$

$\displaystyle \Rightarrow \log \frac{a-b}{2} = (\log a^{\frac{1}{2}} + \log b^{\frac{1}{2}})$

$\displaystyle \Rightarrow \log \frac{a-b}{2} = (\log (ab)^{\frac{1}{2}})$

$\displaystyle \Rightarrow \frac{a-b}{2} = (ab)^{\frac{1}{2}}$

$\displaystyle \Rightarrow a^2 + b^2 -2ab = 4ab$

$\displaystyle \Rightarrow a^2 + b^2 = 6ab$

$\displaystyle \\$

Question 13: $\displaystyle \text{If } a^2 + b^2 = 23 ab \text{ , prove that } \log \Big( \frac{a+b}{5} \Big) = \frac{1}{2} (\log a + \log b)$

$\displaystyle a^2 + b^2 = 23 ab$

$\displaystyle \Rightarrow a^2+b^2 + 2ab = 25 ab$

$\displaystyle \Rightarrow (a+b)^2 = 25ab$

$\displaystyle \Rightarrow \log (a+b)^2 = \log 5^2ab$

$\displaystyle \Rightarrow 2 \log (a+b) = 2 \log 5 + \log a + \log b$

$\displaystyle \Rightarrow 2 \log (a+b) - 2\log 5 = \log a + \log b$

$\displaystyle \Rightarrow \log \frac{a+b}{5} = \frac{1}{2} (\log a + \log b)$

$\displaystyle \\$

Question 14: $\displaystyle \text{If } y = \log_{10} x$, find the following in terms of $\displaystyle y$

$\displaystyle \text{(i) } y = \log_{10} x \Rightarrow 10^y = x$

$\displaystyle \text{(ii) } \log_{10} \sqrt[3]{x^2} = \log_{10} (x^2)^{1/3} = \frac{2}{3} \log_{10} x = \frac{2}{3} y$

$\displaystyle \\$

Question 15: $\displaystyle \text{If } x = \log \frac{2}{3}$ , $\displaystyle y = \log \frac{3}{5} \text{ and } z =2 \log \sqrt{\frac{5}{2}}$ , find the value of $\displaystyle x + y + z \text{ and } 5^{x+y+z}$

$\displaystyle \text{Given } x = \log \frac{2}{3}$ , $\displaystyle y = \log \frac{3}{5} \text{ and } z = 2\log \sqrt{\frac{5}{2}}$

$\displaystyle \text{Therefore } x + y + z = \log \frac{2}{3} + \log \frac{3}{5} + 2 \log \sqrt{\frac{5}{2}} = \log \Big( \frac{2}{3} \times \frac{3}{5} \times \frac{5}{2} \Big) = \log 1 = 0$

$\displaystyle \text{Therefore } 5^{x+y+z} = 5^0 = 1$

$\displaystyle \\$

Question 16: $\displaystyle \text{If } x = \log \frac{3}{5} , y = \log \frac{5}{4} \text{ and } z = 2\log \frac{\sqrt{3}}{2}$ , find the value of $\displaystyle x + y - z \text{ and } 7^{x+y-z}$

$\displaystyle \text{Given } x = \log \frac{3}{5} , y = \log \frac{5}{4} \text{ and } z = 2\log \frac{\sqrt{3}}{2}$

$\displaystyle x + y - z = \log \frac{3}{5} + \log \frac{5}{4} - 2 \log \frac{\sqrt{3}}{2} = \log \Big( \frac{3}{5} \times \frac{5}{4} \times \frac{4}{3} \Big) = \log 1 = 0$

$\displaystyle \text{Therefore } 7^{x+y-z} = 7^0 = 1$

$\displaystyle \\$

Question 17: $\displaystyle \text{If } x^2 = \log_{10} a, y^3 = \log_{10} b \text{ and } \frac{x^2}{2} - \frac{y^3}{3} = \log_{10} c$, express $\displaystyle c$ in terms of $\displaystyle a \text{ and } b$.

$\displaystyle \text{Given } \log_{10} c = \frac{x^2}{2} - \frac{y^3}{3}$

$\displaystyle \log_{10} c = \frac{1}{2} \log_{10} a - \frac{1}{3} \log_{10} b$

$\displaystyle = \log_{10} a^{\frac{1}{2}} - \log_{10} b^{\frac{1}{3}}$

$\displaystyle = \log_{10} \frac{a^{\frac{1}{2}}}{b^{\frac{1}{3}}}$

$\displaystyle c = \frac{a^{\frac{1}{2}}}{b^{\frac{1}{3}}}$

$\displaystyle \\$

Question 18: $\displaystyle \text{If } 2 \log_{10} a + \log_{10} b = 2$, express $\displaystyle b$ in terms of $\displaystyle a$

$\displaystyle 2 \log_{10} a + \log_{10} b = 2$

$\displaystyle 2 \log_{10} a + \log_{10} b = \log_{10} 100$

$\displaystyle \log_{10} b = \log_{10} 100 -\log_{10} a^2$

$\displaystyle \log_{10} b = \log_{10} \frac{100}{a^2}$

$\displaystyle \Rightarrow b = \frac{100}{a^2}$

$\displaystyle \\$

Question 19: $\displaystyle \text{If } \log x = a + b \text{ and } \log y = a - b$, express $\displaystyle \log \Big( \frac{10x}{y^2} \Big)$ in terms of $\displaystyle a \text{ and } b$.

$\displaystyle \text{Given } \log x = a + b \text{ and } \log y = a - b$

$\displaystyle \log \Big( \frac{10x}{y^2} \Big) = \log 10x - \log y^2$

$\displaystyle = \log 10 + \log x - 2 \log y$

$\displaystyle = 1 + a + b -2(a-b)$

$\displaystyle = 3b-a+1$

$\displaystyle \\$

Question 20: $\displaystyle \text{If } \log x = a + b \text{ and } \log y = a - b$, express $\displaystyle \log x^2y$ in terms of $\displaystyle a \text{ and } b$.

Given: $\displaystyle \log x = a + b \text{ and } \log y = a - b$

$\displaystyle \log x^2y = \log x^2 + \log y$

$\displaystyle = 2 \log x + \log y$

$\displaystyle = 2 (a+b) + a-b$

$\displaystyle = 3a+b$

$\displaystyle \\$

Question 21: Solve the following:

$\displaystyle \text{(i) } \log (x+1) - \log (x-1) = 1 \Rightarrow \log \frac{x+1}{x-1} = \log 10$

$\displaystyle \text{(ii) } \log (2x+1) - \log (2x-1) = 1 \Rightarrow \log \frac{2x+1}{2x-1} = \log 10$

$\displaystyle \text{(iii) } 3^{\log x} - 2^{\log x} = 2^{\log x+1}-3^{\log x-1}$

$\displaystyle \text{(iv) } \log_2 x + \log_4 x + \log_{64} x = 5$

$\displaystyle \text{(viii) } \log (x+1) + \log (x-1) = 3 \log 2 + \log 3, x >0$

$\displaystyle \text{(ix) } \log_3 x + \log_9 x + \log_{81} x = \frac{7}{4}$

$\displaystyle \text{(x) } \log_2 x + \log_8 x + \log_{32} x = \frac{23}{15}$

$\displaystyle \text{(i) } \log (x+1) - \log (x-1) = 1 \Rightarrow \log \frac{x+1}{x-1} = \log 10$

$\displaystyle \Rightarrow \frac{x+1}{x-1} = 10 \Rightarrow x+1 = 10x-10 \Rightarrow x = \frac{11}{9}$

$\displaystyle \text{(ii) } \log (2x+1) - \log (2x-1) = 1 \Rightarrow \log \frac{2x+1}{2x-1} = \log 10$

$\displaystyle \Rightarrow \frac{2x+1}{2x-1} = 10 \Rightarrow 2x+1 = 20x - 10 \Rightarrow x = \frac{11}{18}$

$\displaystyle \text{(iii) } 3^{\log x} - 2^{\log x} = 2^{\log x+1}-3^{\log x-1}$

$\displaystyle \Rightarrow 3^{\log x} + \frac{3^{\log x}}{3} = 2.2^{\log x} + 2^{\log x}$

$\displaystyle \Rightarrow \frac{4}{3} \times 3^{\log x} = 3 \times 2^{\log x}$

$\displaystyle \Rightarrow \frac{3^{\log x}}{2^{\log x}} = \frac{9}{4}$

$\displaystyle \Rightarrow \Big( \frac{3}{2} \Big)^{\log x} = \Big( \frac{3}{2} \Big)^2$

$\displaystyle \Rightarrow \log x = 2 \Rightarrow \log x = \log 100 \Rightarrow x = 100$

$\displaystyle \text{(iv) } \log_2 x + \log_4 x + \log_{64} x = 5$

$\displaystyle \Rightarrow \log_2 x + \log_{2^2} x^1 + \log_{2^6} x^1 = 5$

$\displaystyle \Rightarrow \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{6} \log_2 x = 5$

$\displaystyle \Rightarrow (1+ \frac{1}{2} + \frac{1}{6} ) \log_2 x = 5$

$\displaystyle \Rightarrow \frac{10}{6} \log_2 x = 5$

$\displaystyle \Rightarrow \log_2 x = 3 \Rightarrow \log_2 x = \log_2 8 \Rightarrow x = 8$

$\displaystyle \text{(v) } \log (3x+2) + \log (3x-2) = 5 \log 2$

$\displaystyle \Rightarrow \log (3x+2) (3x-2) = \log 2^5$

$\displaystyle \Rightarrow 9x^2 - 4 = 32$

$\displaystyle \Rightarrow 9x^2= 36$

$\displaystyle \Rightarrow x^2 = 4$

$\displaystyle \Rightarrow x = 2$

$\displaystyle \text{(vi) } \log_3 (x+1) - 1 = 3 + \log_3 (x-1)$

$\displaystyle \Rightarrow \log_3 (x+1) - \log_3 3 = \log_3 3^3 + \log_3 (x-1)$

$\displaystyle \Rightarrow \log_3 (x+1) - \log_3 (x-1) = \log_3 3 + \log_3 3^3$

$\displaystyle \Rightarrow \log_3 \Big( \frac{x+1}{x-1} \Big) = \log_3 (3 \times 3^3)$

$\displaystyle \Rightarrow \frac{x+1}{x-1} = 81$

$\displaystyle \Rightarrow x+1 = 81x - 81$

$\displaystyle \Rightarrow or \ x = \frac{82}{80} = \frac{41}{40}$

$\displaystyle \text{(vii) } \log_x 25 - \log_x 5 + \log_x \Big( \frac{1}{125} \Big) =2$

$\displaystyle \Rightarrow 2 \log_x 5 - \log_x 5 + \log_x (5^{-3}) = 2$

$\displaystyle \Rightarrow \log_x 5 - 3 \log_x 5 = \log_x x^2$

$\displaystyle \Rightarrow -2 \log_x 5 = \log_x x^2$

$\displaystyle \Rightarrow x^2 = 5^{-2}$

$\displaystyle \Rightarrow x^2 = ( \frac{1}{5} )^2$

$\displaystyle \text{(viii) } \log (x+1) + \log (x-1) = 3 \log 2 + \log 3, x >0$

$\displaystyle \Rightarrow \log (x^2 -1) = \log(2^3 \times 3)$

$\displaystyle \Rightarrow x^2 -1 = 24$

$\displaystyle \Rightarrow x^2 = 25$

$\displaystyle \Rightarrow x = 5$

$\displaystyle \text{(ix) } \log_3 x + \log_9 x + \log_{81} x = \frac{7}{4}$

$\displaystyle \Rightarrow \log_3 x + \log_{3^2} x^1 + \log_{3^4} x^1 = \frac{7}{4}$

$\displaystyle \Rightarrow \log_3 x + \frac{1}{2} \log_3 x + \frac{1}{4} \log_3 x = \frac{7}{4}$

$\displaystyle \Rightarrow \log_3 x ( \frac{4+2+1}{4} ) = \frac{7}{4}$

$\displaystyle \Rightarrow \log_3 x = 1 = \log_3 3$

$\displaystyle \Rightarrow x = 3$

$\displaystyle \text{(x) } \log_2 x + \log_8 x + \log_{32} x = \frac{23}{15}$

$\displaystyle \Rightarrow \log_2 x + \log_{2^3} x^1 + \log_{2^5} x^1 = \frac{23}{15}$

$\displaystyle \Rightarrow \log_2 x + \frac{1}{3} \log_2 x + \frac{1}{5} \log_2 x = \frac{23}{15}$

$\displaystyle \Rightarrow \log_2 x \Big( 1 + \frac{1}{3} + \frac{1}{5} \Big) = \frac{23}{15}$

$\displaystyle \Rightarrow \log_2 x ( \frac{23}{15} ) = \frac{23}{15}$

$\displaystyle \Rightarrow \log_2 x = 1 = \log_2 2$

$\displaystyle \Rightarrow x = 2$

$\displaystyle \\$

Question 22: $\displaystyle \text{If } \frac{\log x}{\log 5} = \frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}}$ , find $\displaystyle x \text{ and } y$

$\displaystyle \text{Given } \frac{\log x}{\log 5} = \frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}}$

$\displaystyle \Rightarrow \frac{\log x}{\log 5} = \frac{\log 9}{\log \frac{1}{3}} = \frac{\log 3^2}{\log 3^{-1}} = -2$

$\displaystyle \Rightarrow \log x = -2 \log 5 = \log 5^{-2} \Rightarrow x = \frac{1}{25}$

$\displaystyle \frac{\log y^2}{\log 2} = \frac{\log 9}{\log \frac{1}{3}}$

$\displaystyle \Rightarrow \frac{\log y^2}{\log 2} = \log 2 \frac{ \log 9}{\log 3^{-1}}$

$\displaystyle \Rightarrow \frac{\log y^2}{\log 2} = \log 2 \frac{2 \log 3}{-1 . \log 3} = -2 \log 2 = \log \frac{1}{4}$

$\displaystyle \Rightarrow y^2 = \frac{1}{4} \Rightarrow y = \frac{1}{2}$

$\displaystyle \\$

Question 23: $\displaystyle \text{If } a = \log_{10} 20 \text{ and } b = \log_{10} 25$, find the value of $\displaystyle x$, $\displaystyle \text{If } 2 \log_{10}(x+1)=2a-b$

$\displaystyle \text{Given } 2a-b = 2 \log_{10}20 - \log_{10} 25 = \log_{10} \Big( \frac{20 \times 20}{25} \Big) = \log_{10} 4^2$

$\displaystyle 2 \log_{10} (x+1) = (2a-b)$

$\displaystyle \log_{10} (x+1)^2 = \log_{10} 4^2$

$\displaystyle \text{Therefore } (x+1)^2 = 4^2 \Rightarrow x + 1 = 4 \Rightarrow x = 3$

$\displaystyle \\$

Question 24: $\displaystyle \text{Given } \log 2 = 0.3010 \text{ and } \log 3 = 0.4771$, find the value of each of the following:

$\displaystyle \text{(i) } \log 12 = \log (3 \times 4) = \log 3 + 2 \log 2 = 0.4771 + 2 \times 0.3010 = 1.0791$

$\displaystyle \text{(ii) } \log 5 = \log \frac{10}{2} = \log 10 - \log 2 = 1 - 0.3010 = 0.6990$

$\displaystyle \text{(iii) } \log 5^{\frac{1}{3}} = \frac{1}{3} (\log 10 - \log 2 ) = \frac{1}{3} (1 - 0.3010 ) = 0.2329$

$\displaystyle \text{(iv) } \log 108 = \log 27 + \log 4 = 3 \log 3 + 2 \log 2 = 3 \times 0.4771 + 2 \times 0.3010 = 2.034$

$\displaystyle \text{(v) } \log \Big( \frac{3}{8} \Big) = \log 3 - \log 8 = \log 3 - 2 \log 2 = 0.4771 - 3(0.3020) = -0.4259$

$\displaystyle \text{(vi) } \log 48 = \log 16 + \log 3 = 4 \log 2 + \log 3 = 4(0.3010) + 0.4771 = 1.6812$

$\displaystyle \\$

Question 25: Without using $\displaystyle \log$ tables show that

$\displaystyle \text{(i) } \frac{\log \sqrt{27} + \log \sqrt{8} - \log \sqrt{125}}{\log 6 - \log 5} = \frac{3}{2}$

$\displaystyle \text{LHS } = \frac{\log \sqrt{27} + \log \sqrt{8} - \log \sqrt{125}}{\log 6 - \log 5}$

$\displaystyle = \frac{\frac{3}{2} \log 3 + \frac{3}{2} \log 2 - \frac{3}{2} \log 5}{\log 6 - \log 5}$

$\displaystyle = \frac{3}{2} \frac{ \log \frac{3 \times 2}{ 5 } }{\log 6 - \log 5}$

$\displaystyle = \frac{3}{2} =$ RHS

$\displaystyle \text{(ii) } \frac{\log \sqrt{27} + \log 8 - \log \sqrt{1000}}{\log 1.2 } = \frac{3}{2}$

$\displaystyle = \frac{\frac{3}{2}\log 3 + \frac{3}{2}\log 2 - \frac{3}{2}\log 10 }{\log 1.2 }$

$\displaystyle = \frac{3}{2} (\frac{\log 3 + 2\log 2 - \log 10 }{\log 1.2 })$

$\displaystyle = \frac{3}{2} \log \frac{\frac{3 \times 4}{10} }{ \log 1.2}$

$\displaystyle = \frac{3}{2}$

$\displaystyle \\$