Answer: Option B. -> $$\frac{{1000}}{{301}}$$
$$\eqalign{
& {\log _2}10 \cr
& = \frac{1}{{{{\log }_{10}}2}} \cr
& = \frac{1}{{0.3010}} \cr
& = \frac{{10000}}{{3010}} \cr
& = \frac{{1000}}{{301}} \cr} $$

Answer: Option B. -> 3
$$\eqalign{
& \Rightarrow {\log _{10}}5 + {\log _{10}}\left( {5x + 1} \right) = {\log _{10}}\left( {x + 5} \right) + 1 \cr
& \Rightarrow {\log _{10}}\left[ {5\left( {5x + 1} \right)} \right] = {\log _{10}}\left[ {10\left( {x + 5} \right)} \right] \cr
& \Rightarrow 5\left( {5x + 1} \right) = 10\left( {x + 5} \right) \cr
& \Rightarrow 5x + 1 = 2x + 10 \cr
& \Rightarrow 3x = 9 \cr
& \Rightarrow x = 3 \cr} $$

Answer: Option B. -> 1.9030
$$\eqalign{
& {\log _{10}}80 \cr
& = {\log _{10}}\left( {8 \times 10} \right) \cr
& = {\log _{10}}8 + {\log _{10}}10 \cr
& = {\log _{10}}\left( {{2^3}} \right) + 1 \cr
& = 3{\log _{10}}2 + 1 \cr
& = \left( {3 \times 0.3010} \right) + 1 \cr
& = 1.9030 \cr} $$

Answer: Option B. -> 1
Given expression
$$\eqalign{
& = {\log _{60}}3 + {\log _{60}}4 + {\log _{60}}5 \cr
& = {\log _{60}}\left( {3 \times 4 \times 5} \right) \cr
& = {\log _{60}}60 \cr
& = 1 \cr} $$

Answer: Option B. -> 5
$$\eqalign{
& {\text{lo}}{{\text{g}}_5}\frac{{\left( {125} \right)\left( {625} \right)}}{{25}} \cr
& = {\text{lo}}{{\text{g}}_5}\left( {\frac{{{5^3} \times {5^4}}}{{{5^2}}}} \right) \cr
& = {\text{lo}}{{\text{g}}_5}{5^5} \cr
& {\text{ = 5lo}}{{\text{g}}_5}5 \cr
& = 5 \cr} $$

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

Answer: Option A. -> p = q15
$$\eqalign{
& {\text{lo}}{{\text{g}}_8}{\text{p}} = 25\,\,{\text{and lo}}{{\text{g}}_2}{\text{q}} = 5\, \cr
& \Rightarrow {\text{p = }}{{\text{8}}^{25}}\,{\text{and q}} = {2^5} \cr
& \Rightarrow {\text{p = }}{\left( {{2^3}} \right)^{25}}{\text{ and q = }}{2^5} \cr
& \Rightarrow {\text{p = }}{{\text{2}}^{75}}{\text{ and q = }}{2^5} \cr
& \Rightarrow {\text{p = }}{\left( {{2^5}} \right)^{15}}{\text{ and q = }}{2^5} \cr
& \Rightarrow {\text{ p = }}{{\text{q}}^{15}} \cr} $$

Answer: Option A. -> 1
$$\eqalign{
& {\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {{\text{5lo}}{{\text{g}}_{10}}100} \right)\,} \right]^2} \cr
& = {\left[ {{\text{lo}}{{\text{g}}_{10}}\left\{ {{\text{5lo}}{{\text{g}}_{10}}{{\left( {10} \right)}^2}} \right\}\,} \right]^2} \cr
& = \,{\left[ {{\text{lo}}{{\text{g}}_{10}}\left( {5 \times 2} \right)} \right]^2} \cr
& = {\left( {{\text{lo}}{{\text{g}}_{10}}10} \right)^2} \cr
& = 1 \cr} $$

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

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