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

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

Answer: Option D. -> 3
$$\eqalign{
& {\log _{10}}125 + {\log _{10}}8 = x \cr
& \Rightarrow {\log _{10}}\left( {125 \times 8} \right) = x \cr
& \Rightarrow x = {\log _{10}}\left( {1000} \right) \cr
& \Rightarrow x = {\log _{10}}{\left( {10} \right)^3} \cr
& \Rightarrow x = 3{\log _{10}}10 \cr
& \Rightarrow x = 3 \cr} $$

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

Answer: Option B. -> $$\frac{{2b}}{3}$$
$$\eqalign{
& a = {\log _8}225 \cr
& = {\log _{{2^3}}}\left( {{{15}^2}} \right) \cr
& = \frac{2}{3}{\log _2}15 \cr
& = \frac{{2b}}{3} \cr} $$

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

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

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

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

Answer: Option A. -> - (1 + a)
$$\eqalign{
& {\log _{10}}\left( {{1 \over {70}}} \right) \cr
& = {\log _{10}}1 - {\log _{10}}70 \cr
& = - {\log _{10}}\left( {7 \times 10} \right) \cr
& = - \left( {{{\log }_{10}}7 + {{\log }_{10}}10} \right) \cr
& = - \left( {a + 1} \right) \cr} $$