Question 1: Simplify each of the following:

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

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

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

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

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

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Question 2: Prove the following

(i) $\log 108 = 3 \log 3 + 2 \log 2$

LHS $= \log 108 = \log (27 \times 4) = \log 27 + \log 4 = 3 \log 3 + 2 \log 2 =$ RHS

(ii) $\log 75 = \log 3 + 2 \log 5$

LHS $= \log 75 = \log (25 \times 3) = \log 25 + \log 3 = 2 \log 5 + \log 3 =$ RHS

(iii) $\log 72 = 2 \log 3 + 3 \log 2$

LHS $= \log (8 \times 9) = \log 8 + \log 9 = 3 \log 2 + 2 \log 3 =$ RHS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence LHS = RHS

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

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

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

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

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

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

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

(i) $\log 3 + 2 = \log 3 + \log 100 = \log 300$

(ii) $2 \log 3 + 3 = \log 3^2 + \log 1000 = \log 9000$

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

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

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

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

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

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

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Question 4: Evaluate the following:

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

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

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

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

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

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

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

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

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Question 5: Prove that:

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

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

RHS $= \log_3 3^2 = 2 \log_3 3 = 2$. Hence proved.

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

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

RHS $= \log_5 25 = \log_5 5^2 = 2 \log_5 5 = 2$. Hence proved.

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

Answer:

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

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

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

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

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

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

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

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

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

Answer:

$2 \log x$ and $2 \log y -2 = 0$

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

$\Rightarrow a^2b = 100$

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

Therefore $5^{x+y+z} = 5^0 = 1$

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

Answer:

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

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

Therefore $7^{x+y-z} = 7^0 = 1$

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

Given $\log x = a + b$ and $\log y = a - b$

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

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

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

$= 3b-a+1$

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

Answer:

Given: $\log x = a + b$ and $\log y = a - b$

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

$= 2 \log x + \log y$

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

$= 3a+b$

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Question 21: Solve the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(v) $\log (3x+2) + \log (3x-2) = 5 \log 2$

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

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

$\Rightarrow 9x^2= 36$

$\Rightarrow x^2 = 4$

$\Rightarrow x = 2$

(vi) $\log_3 (x+1) - 1 = 3 + \log_3 (x-1)$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow x^2 -1 = 24$

$\Rightarrow x^2 = 25$

$\Rightarrow x = 5$

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

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

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

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

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

$\Rightarrow x = 3$

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

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

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

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

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

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

$\Rightarrow x = 2$

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

Answer:

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

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

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

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

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

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

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

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

Answer:

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

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

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

Therefore $(x+1)^2 = 4^2 \Rightarrow x + 1 = 4 \Rightarrow x = 3$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$= \frac{3}{2}$

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