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

$$\eqalign{ & {\log _2}10 \cr & = \frac{1}{{{{\log }_{10}}2}} \cr & = \frac{1}{{0.3010}} \cr & = \frac{{10000}}{{3010}} \cr & = \frac{{1000}}{{301}} \cr} $$

$$\eqalign{ & {\log _{10}}80 \cr & = {\log _{10}}\left( {8 \times 10} \right) \cr & = {\log _{10}}8 + {\log _{10}}10 \cr & = {\log _{10}}\left( {{2^3}} \right) + 1 \cr & = 3{\log _{10}}2 + 1 \cr & = \left( {3 \times 0.3010} \right) + 1 \cr & = 1.9030 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_5}\frac{{\left( {125} \right)\left( {625} \right)}}{{25}} \cr & = {\text{lo}}{{\text{g}}_5}\left( {\frac{{{5^3} \times {5^4}}}{{{5^2}}}} \right) \cr & = {\text{lo}}{{\text{g}}_5}{5^5} \cr & {\text{ = 5lo}}{{\text{g}}_5}5 \cr & = 5 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_{3\sqrt 2 }}\left( {\frac{1}{{18}}} \right)\, \cr & = {\text{lo}}{{\text{g}}_{3\sqrt 2 }}\left( {\frac{1}{{{{\left( {3\sqrt 2 } \right)}^2}}}} \right)\, \cr & = {\text{lo}}{{\text{g}}_{3\sqrt 2 }}{\text{ }}{\left( {3\sqrt 2 } \right)^{ - 2}} \cr & = \left( { - 2} \right){\text{lo}}{{\text{g}}_{3\sqrt 2 }}3\sqrt 2 \cr & = - 2 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_{10}}\left( {0.0001} \right)\, \cr & = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{10000}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}\left( {\frac{1}{{{{10}^4}}}} \right) \cr & \, = {\text{lo}}{{\text{g}}_{10}}{10^{ - 4}} \cr & \, = - 4\,{\text{lo}}{{\text{g}}_{10}}10 \cr & = - 4{\text{ }} \cr} $$

$$\eqalign{ & \log \left( {{2^{64}}} \right) \cr & = 64 \times \log 2 \cr & = \left( {64 \times 0.30103} \right) \cr & = 19.26592 \cr} $$
Its characteristic is 19.
Hence, then number of digits in 2⁶⁴ is 20.

$$\eqalign{ & {\log _x}\left( {{9 \over {16}}} \right) = - {1 \over 2} \cr & \Rightarrow {x^{ - {1 \over 2}}} = {9 \over {16}} \cr & \Rightarrow {1 \over {\sqrt x }} = {9 \over {16}} \cr & \Rightarrow \sqrt x = {{16} \over 9} \cr & \Rightarrow x = {\left( {{{16} \over 9}} \right)^2} \cr & \Rightarrow x = {{256} \over {81}} \cr} $$

$$\eqalign{ & {\text{Given}}\,\,\,{\text{Expression}} \cr & = \frac{1}{{{{\log }_a}bc + {{\log }_a}a}} + \frac{1}{{{{\log }_b}ca + {{\log }_b}b}} + \frac{1}{{{{\log }_c}ab + {{\log }_c}c}} \cr & = \frac{1}{{{{\log }_a}\left( {abc} \right)}} + \frac{1}{{{{\log }_b}\left( {abc} \right)}} + \frac{1}{{{{\log }_c}\left( {abc} \right)}} \cr & = {\log _{abc}}a + {\log _{abc}}b + {\log _{abc}}c \cr & = {\log _{abc}}\left( {abc} \right) \cr & = 1 \cr} $$

$$\eqalign{ & {\log _5}12 = {\log _5}\left( {3 \times 4} \right) \cr & = {\log _5}3 + {\log _5}4 \cr & = {\log _5}3 + 2{\log _5}2 \cr & = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}5}} + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}5}} \cr} $$
$$ = \frac{{{{\log }_{10}}3}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$   $$ + \frac{{2{{\log }_{10}}2}}{{{{\log }_{10}}10 - {{\log }_{10}}2}}$$
$$\eqalign{ & = \frac{b}{{1 - a}} + \frac{{2a}}{{1 - a}} \cr & = \frac{{2a + b}}{{1 - a}} \cr} $$

$$\eqalign{ & {\log _{10}}125 + {\log _{10}}8 = x \cr & \Rightarrow {\log _{10}}\left( {125 \times 8} \right) = x \cr & \Rightarrow x = {\log _{10}}\left( {1000} \right) \cr & \Rightarrow x = {\log _{10}}{\left( {10} \right)^3} \cr & \Rightarrow x = 3{\log _{10}}10 \cr & \Rightarrow x = 3 \cr} $$

$$\eqalign{ & {\log _5}\left( {{x^2} + x} \right) - {\log _5}\left( {x + 1} \right) = 2 \cr & \Rightarrow {\log _5}\left( {\frac{{{x^2} + x}}{{x + 1}}} \right) = 2\, \cr & \Rightarrow {\log _5}\left[ {\frac{{x\left( {x + 1} \right)}}{{x + 1}}} \right] = 2 \cr & \Rightarrow {\log _5}x = 2 \cr & \Rightarrow x = {5^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25 \cr} $$

$$\eqalign{ & \Rightarrow {\log _{10}}5 + {\log _{10}}\left( {5x + 1} \right) = {\log _{10}}\left( {x + 5} \right) + 1 \cr & \Rightarrow {\log _{10}}\left[ {5\left( {5x + 1} \right)} \right] = {\log _{10}}\left[ {10\left( {x + 5} \right)} \right] \cr & \Rightarrow 5\left( {5x + 1} \right) = 10\left( {x + 5} \right) \cr & \Rightarrow 5x + 1 = 2x + 10 \cr & \Rightarrow 3x = 9 \cr & \Rightarrow x = 3 \cr} $$

Given expression
$$\eqalign{ & = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr & = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr & = {\log _{60}}60 \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{6{\text{ }}{{\log }_{10}}1000}}{{3{\text{ }}{{\log }_{10}}100}} \cr & = \frac{{6{\text{ }}{{\log }_{10}}{{10}^3}}}{{3{\text{ }}{{\log }_{10}}{{10}^2}}} \cr & = \frac{{6 \times 3{\text{ }}{{\log }_{10}}10}}{{3 \times 2{\text{ }}{{\log }_{10}}10}} \cr & = \frac{{18}}{6} \cr & = 3 \cr} $$

$$\eqalign{ & {{\log }_2}\left[ {{\text{ }}{{\log }_3}\left( {{\text{ }}{{\log }_2}x} \right)} \right] = 1 \cr & \Rightarrow {\log _3}\left( {{{\log }_2}x} \right) = {2^1} = 2 \cr & \Rightarrow {\log _2}x = {3^2} = 9 \cr & \Rightarrow x = {2^9} = 512 \cr} $$

$$\eqalign{ & \log \left( {\frac{9}{{14}}} \right) - \log \left( {\frac{{15}}{{16}}} \right) + \log \left( {\frac{{35}}{{24}}} \right) \cr & = \log \left( {\frac{9}{{14}} \div \frac{{15}}{{16}} \times \frac{{35}}{{24}}} \right) \cr & = \log \left( {\frac{9}{{14}} \times \frac{{16}}{{15}} \times \frac{{35}}{{24}}} \right) \cr & = \log 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {{\text{5lo}}{{\text{g}}_{10}}100} \right)\,} \right]^2} \cr & = {\left[ {{\text{lo}}{{\text{g}}_{10}}\left\{ {{\text{5lo}}{{\text{g}}_{10}}{{\left( {10} \right)}^2}} \right\}\,} \right]^2} \cr & = \,{\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {5 \times 2} \right)} \right]^2} \cr & = {\left( {{\text{lo}}{{\text{g}}_{10}}10} \right)^2} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{lo}}{{\text{g}}_8}{\text{p}} = 25\,\,{\text{and lo}}{{\text{g}}_2}{\text{q}} = 5\, \cr & \Rightarrow {\text{p = }}{{\text{8}}^{25}}\,{\text{and q}} = {2^5} \cr & \Rightarrow {\text{p = }}{\left( {{2^3}} \right)^{25}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{p = }}{{\text{2}}^{75}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{p = }}{\left( {{2^5}} \right)^{15}}{\text{ and q = }}{2^5} \cr & \Rightarrow {\text{ p = }}{{\text{q}}^{15}} \cr} $$

$$\eqalign{ & {a^x} = {b^y} \cr & \Rightarrow \log {a^x} = \log {b^y} \cr & \Rightarrow x\log a = y\log b \cr & \Rightarrow {{\log a} \over {\log b}} = {y \over x} \cr} $$