In arithmetic progression common ratio are equal to
$$\log \left( {{3^x} - 2} \right) - \log 3 = $$     $$\log \left( {{3^x} + 4} \right) - $$   $$\log \left( {{3^x} - 2} \right)$$
$$\frac{{\log \left( {{3^x} - 2} \right)}}{{\log 3}} = \frac{{\log \left( {{3^x} + 4} \right)}}{{\log \left( {{3^x} - 2} \right)}}$$     $$\left( {\therefore \log a - \log b = \log \frac{a}{b}} \right)$$
$$\eqalign{ & \frac{{\log {3^x}}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{{x\log 3}}{{\log 2\log 3}} = \frac{{x\log 3\log 4\log 2}}{{x\log 3}} \cr & \frac{x}{{\log 2}} = \log 4\log 2 \cr & x = \log 4\log \,2\log 2 \cr & x = \log 8 \cr & x = \log {2^3} \cr} $$

$$\eqalign{ & a = {\log _8}225 \cr & = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr & = \frac{2}{3}{\log _2}15 \cr & = \frac{{2b}}{3} \cr} $$

$$\eqalign{ & {\text{Given}}, \cr & {\log _{10}}a = p,\,{\log _{10}}b = q \cr & {\log _{10}}\left( {{a^p}{b^q}} \right) = {\log _{10}}{a^p} + {\log _{10}}{b^q} \cr & = p{\log _{10}}a + q{\log _{10}}b \cr & = {p^2} + {q^2} \cr} $$

$$\eqalign{ & {\log _5}512 \cr & = \frac{{\log 512}}{{\log 5}} \cr & = \frac{{\log {2^9}}}{{\log \left( {\frac{{10}}{2}} \right)}} \cr & = \frac{{9\log 2}}{{\log 10 - \log 2}} \cr & = \frac{{\left( {9 \times 0.3010} \right)}}{{1 - 0.3010}} \cr & = \frac{{2.709}}{{0.699}} \cr & = \frac{{2709}}{{699}} \cr & = 3.876 \cr} $$

$$\eqalign{ & {\text{let }}\log x = - 3.153 \cr & {\text{then, }}\log x = - 3.153 \cr & = - 3 + \left( { - 0.153} \right) \cr & = \left( { - 3 - 1} \right) + \left( {1 - 0.153} \right) \cr & = - 4 + 0.847 = \overline 4 .847 \cr & {\text{Hence,}} \cr & {\text{characteristic = - 4, mantissa = 0}}{\text{.847 }} \cr} $$

$$\eqalign{ & {\log _{10}}\left( {\frac{1}{{70}}} \right) \cr & = {\log _{10}}1 - {\log _{10}}70 \cr & = - {\log _{10}}\left( {7 \times 10} \right) \cr & = - \left( {{{\log }_{10}}7 + {{\log }_{10}}10} \right) \cr & = - \left( {a + 1} \right) \cr} $$

$$\eqalign{ & \log x - 5\log 3 = - 2 \cr & \Rightarrow \log x - \log {3^5} = - 2 \cr & \Rightarrow \log \left( {\frac{x}{{{3^5}}}} \right) = - 2 \cr & \Rightarrow \frac{x}{{243}} = {10^{ - 2}} = \frac{1}{{100}} \cr & \Rightarrow x = \frac{{243}}{{100}} = 2.43 \cr} $$

$$\eqalign{ & \log {4^{50}} \cr & = 50\log 4 \cr & = 50\log {2^2} \cr & = \left( {50 \times 2} \right)\log 2 \cr & = 100 \times \log 2 \cr & = \left( {100 \times 0.30103} \right) \cr & = 30.103 \cr & \therefore {\text{characteristic}} = 30, \cr} $$
Hence, the number of digits in $${{\text{4}}^{50}} = 31$$

$$\eqalign{ & \log \left( {{4^9} \times {5^{17}}} \right) \cr & = \log \left( {{4^9}} \right) + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^2}} \right)^9 + \log \left( {{5^{17}}} \right) \cr & = \log \left( {{2^{18}}} \right) + \log \left( {{5^{17}}} \right) \cr & = 18\log 2 + 17\log 5 \cr & = 18\log 2 + 17\left( {\log 10 - \log 2} \right) \cr & = 18\log 2 + 17\log 10 - 17\log 2 \cr & = \log 2 + 17\log 10 \cr & = 0.3010 + 17 \times 1 = 17.3010 \cr & \therefore {\text{Characteristic}} = 17 \cr & {\text{Hence, the number of digits in }}\left( {{4^9} \times {5^{17}}} \right) = 18 \cr} $$

$$\eqalign{ & \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr & \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr & \Rightarrow {\left( {x - y} \right)^2} = 0 \cr & \Rightarrow x - y = 0 \cr & \Rightarrow x = y\, \cr} $$

$$\eqalign{ & \log \frac{a}{b} + \log \frac{b}{a} = \log \left( {a + b} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log \left( {\frac{a}{b} \times \frac{b}{a}} \right) \cr & \Rightarrow \log \left( {a + b} \right) = \log 1 \cr & So,\,\,\,a + b = 1 \cr} $$

$$\eqalign{ & a = {b^2} = {c^3} = {d^4} \cr & \Rightarrow b = {a^{\frac{1}{2}}},\,\,\,\,c = {a^{\frac{1}{3}}},\,\,\,\,d = {a^{\frac{1}{4}}} \cr & \therefore {\log _a}\left( {abcd} \right) \cr & = {\log _a}\left( {a \times {a^{\frac{1}{2}}} \times {a^{\frac{1}{3}}} \times {a^{\frac{1}{4}}}} \right) \cr & = {\log _a}{a^{\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right)}} \cr & = \left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right){\log _a}a \cr & = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \cr} $$

$$\eqalign{ & \Rightarrow {\log _3}x + {\log _9}{x^2} + {\log _{27}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + {\log _{{3^2}}}{x^2} + {\log _{{3^3}}}{x^3} = 9 \cr & \Rightarrow {\log _3}x + \frac{2}{2}{\log _3}x + \frac{3}{3}{\log _3}x = 9 \cr & \Rightarrow 3{\log _3}x = 9 \cr & \Rightarrow {\log _3}x = 3 \cr & \Rightarrow x = {3^3} = 27 \cr} $$

$$\eqalign{ & \,{\log _7}{\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = 0 \cr & \Rightarrow {\log _5}\left( {\sqrt {x + 5} + \sqrt x } \right) = {7^0} = 1 \cr & \Rightarrow \sqrt {x + 5} + \sqrt x = {5^1} = 5 \cr & \Rightarrow {\left( {\sqrt {x + 5} + \sqrt x } \right)^2} = 25 \cr & \Rightarrow \left( {x + 5} \right) + x + 2\sqrt {x + 5} \sqrt x = 25 \cr & \Rightarrow 2x + 2\sqrt{ x}\,\, \sqrt {x + 5} = 20 \cr & \Rightarrow \sqrt {x}\,\, \sqrt {x + 5} = 10 - x \cr & \Rightarrow x\left( {x + 5} \right) = {\left( {10 - x} \right)^2} \cr & \Rightarrow {x^2} + 5x = 100 + {x^2} - 20x \cr & \Rightarrow 25x = 100 \cr & \Rightarrow x = 4 \cr} $$

$$\eqalign{ & {\text{let }}{\log _{10000}}x = - \frac{1}{4} \cr & \Rightarrow x = {\left( {10000} \right)^{ - \frac{1}{4}}} \cr & = {\left( {{{10}^4}} \right)^{ - \frac{1}{4}}} \cr & = {10^{ - 1}} \cr & = \frac{1}{{10}} \cr} $$

$$\eqalign{ & \frac{{\log \sqrt 8 }}{{\log 8}} \cr & = \frac{{\log {{\left( 8 \right)}^{\frac{1}{2}}}}}{{\log 8}} \cr & = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression }} \cr & = {\text{ }}\log \left( {\frac{{{a^2}}}{{bc}} \times \frac{{{b^2}}}{{ac}} \times \frac{{{c^2}}}{{ab}}} \right) \cr & = \log 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & \left( {\frac{{\log a}}{{\log b}} \times \frac{{\log b}}{{\log c}} \times \frac{{\log c}}{{\log a}}} \right) \cr & = 1 \cr} $$

$$\eqalign{ & 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr & = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr & = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr & = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr & = {\log _{10}}100 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_a}\left( {ab} \right) = x \cr & \Rightarrow \frac{{\log ab}}{{\log a}} = x \cr & \Rightarrow \frac{{\log a + \log b}}{{\log a}} = x \cr & \Rightarrow 1 + \frac{{\log b}}{{\log a}} = x \cr & \Rightarrow \frac{{\log b}}{{\log a}} = x - 1 \cr & \Rightarrow \frac{{\log a}}{{\log b}} = \frac{1}{{x - 1}} \cr & \Rightarrow 1 + \frac{{\log a}}{{\log b}} = 1 + \frac{1}{{x - 1}} \cr & \Rightarrow \frac{{\log b}}{{\log b}} + \frac{{\log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{\log b + \log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{{\text{log}}\left( {ab} \right)}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow {\text{lo}}{{\text{g}}_b}\left( {ab} \right) = \frac{x}{{x - 1}} \cr} $$