3.0 Sample Questions

Question 1. Solve

(i). ${\log _{10}}1$

(ii). ${\log _{10}}10$

(iii). ${\log _{10}}100$

(iv). ${\log _{10}}10000$

From property 1,$${\log _{10}}1 = 0$$

From property 2, $${\log _{10}}10 = 1$$

$${\log _{10}}100 = {\log _{10}}{10^2}$$ From property 5, $${\log _{10}}{10^2} = 2{\log _{10}}10$$ From property 2, $$2{\log _{10}}10 = 2 \times 1 = 2$$

$${\log _{10}}10000 = {\log _{10}}{10^4}$$ From property 5, $${\log _{10}}{10^2} = 4{\log _{10}}10$$ From property 2, $$4{\log _{10}}10 = 4 \times 1 = 4$$

Question 2. Solve

${\log _{10}}0.1$

${\log _{10}}0.001$

${\log _{10}}0.00001$

$${\log _{10}}0.1 = {\log _{10}}{10^{-1}}$$ From property 5, $${\log _{10}}{10^{-1}} = -1{\log _{10}}10$$ From property 2, $$(-1){\log _{10}}10 = -1 \times 1 = -1$$

$${\log _{10}}0.001 = {\log _{10}}{10^{-3}}$$ From property 5, $${\log _{10}}{10^{-3}} = -3{\log _{10}}10$$ From property 2, $$(-3){\log _{10}}10 = -3 \times 1 = -3$$

$${\log _{10}}0.00001 = {\log _{10}}{10^{-5}}$$ From property 5, $${\log _{10}}{10^{-5}} = -5{\log _{10}}10$$ From property 2, $$(-5){\log _{10}}10 = -5 \times 1 = -5$$

Question 3. Prove that ${\log _3}{\log _2}{\log _{\sqrt 5 }}(625) = 1$

Solution: $$\begin{equation} \begin{aligned} {\log _3}{\log _2}{\log _{\sqrt 5 }}({5^4}) \\ {\log _3}{\log _2}\left( {2 \times 4{{\log }_5}5} \right)\quad \left( {by,\,{{\log }_{{a^n}}}{b^m} = \frac{m}{n}{{\log }_a}b,\;\;{{\log }_a}a = 1} \right) \\ {\log _3}{\log _2}8 \\ {\log _3}{\log _2}{2^3}\quad \left( {by,\,{{\log }_a}a = 1} \right) \\ {\log _3}3\quad \left( {by,\,{{\log }_a}a = 1} \right) \\ 1 \\\end{aligned} \end{equation} $$

Question 4. Prove that ${\log _3}{\log _2}{\log _{\sqrt 5 }}(625) = 1$

Solution: $$\begin{equation} \begin{aligned} {\log _3}{\log _2}{\log _{\sqrt 5 }}({5^4}) \\ {\log _3}{\log _2}\left( {2 \times 4{{\log }_5}5} \right)\quad \left( {by,\,{{\log }_{{a^n}}}{b^m} = \frac{m}{n}{{\log }_a}b,\;\;{{\log }_a}a = 1} \right) \\ {\log _3}{\log _2}8 \\ {\log _3}{\log _2}{2^3}\quad \left( {by,\,{{\log }_a}a = 1} \right) \\ {\log _3}3\quad \left( {by,\,{{\log }_a}a = 1} \right) \\ 1 \\\end{aligned} \end{equation} $$

Question 5. If ${a^2} + {b^2} = 23ab$, then prove that $\log \frac{{a + b}}{5} = \frac{1}{2}\left( {\log a + \log b} \right)$

Given,$$\begin{equation} \begin{aligned} {a^2} + {b^2} = 23ab\quad or\quad {a^2} + {b^2} + 2ab = 25ab \\ {(a + b)^2} = {5^2}ab \\ {\left( {\frac{{a + b}}{5}} \right)^2} = ab \\\end{aligned} \end{equation} $$Taking log on both sides we get,$$\begin{equation} \begin{aligned} \log {\left( {\frac{{a + b}}{5}} \right)^2} = \log ab \\ 2\log \left( {\frac{{a + b}}{5}} \right) = \log a + \log b \\ \log \left( {\frac{{a + b}}{5}} \right) = \frac{{\left( {\log a + \log b} \right)}}{2} \\\end{aligned} \end{equation} $$

Question 6. Prove that $7\log \frac{{16}}{{15}} + 5\log \frac{{25}}{{24}} + 3\log \frac{{81}}{{80}} = \log 2$

Solution: $$\begin{equation} \begin{aligned} 7\log \frac{{16}}{{15}} + 5\log \frac{{25}}{{24}} + 3\log \frac{{81}}{{80}} \\ 7(\log 16 - \log 15) + 5(\log 25 - \log 24) + 3(\log 81 - \log 80) \\ 7(\log {2^4} - \log (3 \times 5)) + 5(\log {5^2} - \log ({2^3} \times 3)) + 3(\log {3^4} - \log ({2^4} \times 5)) \\ 7(4\log 2 - \log 3 - \log 5) + 5(2\log 5 - 3\log 2 - \log 3) + 3(4\log 3 - 4\log 2 - \log 5) \\ (28 - 15 - 12)\log 2 + ( - 7 - 5 + 12)\log 3 + ( - 7 + 10 - 3)\log 5 \\ \log 2 \\\end{aligned} \end{equation} $$

Question 7. Prove that $\frac{1}{{{{\log }_2}N}} + \frac{1}{{{{\log }_3}N}} + \frac{1}{{{{\log }_4}N}} + ... + \frac{1}{{{{\log }_{1988}}N}} = \frac{1}{{{{\log }_{2016!}}N}}$

Solution: $$\begin{equation} \begin{aligned} \frac{1}{{{{\log }_2}N}} + \frac{1}{{{{\log }_3}N}} + \frac{1}{{{{\log }_4}N}} + ... + \frac{1}{{{{\log }_{2016}}N}} \\ {\log _N}2 + {\log _N}3 + {\log _N}4 + ... + {\log _N}2016 \\ {\log _N}(2.3.4......2016) \\ {\log _N}(1.2.3.4....2016) \\ {\log _N}2016! \\ \frac{1}{{{{\log }_{2016!}}N}} \\\end{aligned} \end{equation} $$

Question 8. If $0<x<1$, prove that, $\log (1 + x) + \log (1 + {x^2}) + \log (1 + {x^4}) + ....\infty = - \log (1 - x)$

Solution: $$\begin{equation} \begin{aligned} \log (1 + x) + \log (1 + {x^2}) + \log (1 + {x^4}) + ....\infty \\ \log [(1 + x).(1 + {x^2}).(1 + {x^4}).....\infty ] \\ \log [1 + x + {x^2} + {x^3} + {x^4} + ...\infty ] \\ \log \frac{1}{{1 - x}}\quad for\quad \left| x \right| < 1 \\ \log 1 - \log (1 - x) \\ - \log (1 - x) \\\end{aligned} \end{equation} $$