Answer: Option B. -> 1

Given expression
= log₆₀ 3 + log₆₀ 4 + log₆₀ 5
= log₆₀ (3 x 4 x 5)
= log₆₀ 60
= 1.

Answer: Option B. -> 3

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1

log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

log₁₀ [5 (5x + 1)] = log₁₀ [10(x + 5)]

5(5x + 1) = 10(x + 5)

5x + 1 = 2x + 10

3x = 9

x = 3.

Answer: Option B. -> log (2 + 3) = log (2 x 3)
$$\eqalign{
& {\text{A = Since lo}}{{\text{g}}_a}a = 1,{\text{ so lo}}{{\text{g}}_{10}}10 = 1 \cr
& \cr
& B = \,{\text{log}}\left( {2 + 3} \right) = 5\, \cr
& \,\,\,{\text{and log}}\left( {2 \times 3} \right) = {\text{log}}6 \cr
& = {\text{log}}2 + {\text{log}}3 \cr
& \therefore \,{\text{log}}\left( {2 + 3} \right) \ne {\text{log}}\left( {2 \times 3} \right). \cr
& \cr
& C = \,{\text{ Since lo}}{{\text{g}}_a}1 = 0,so{\text{ lo}}{{\text{g}}_{10}}1 = 0. \cr
& \cr
& {\text{D = log}}\left( {1 + 2 + 3} \right) = {\text{ log}}6 \cr
& = {\text{ log}}\left( {1 \times 2 \times 3} \right) \cr
& = {\text{ log}}1 + {\text{ log}}2 + {\text{ log}}3. \cr
& {\text{So (B) is incorrect}} \cr} $$

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

Answer: Option C. -> 0.954
$$\eqalign{
& \log 27 = 1.431 \cr
& \Rightarrow \log \left( {{3^3}} \right) = 1.431 \cr
& \Rightarrow 3\,\log \,3 = 1.431 \cr
& \Rightarrow \log \,3 = 0.477 \cr
& \therefore \log \,9 = \log \left( {{3^2}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\,\log \,3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2 \times 0.477} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.954 \cr} $$

Answer: Option C. -> $$\frac{1}{2}$$
$$\eqalign{
& \frac{{\log \sqrt 8 }}{{\log 8}} = {\frac{{\log \left( 8 \right)}}{{\log 8}}^{\frac{1}{2}}} \cr
& = \frac{{\frac{1}{2}\log 8}}{{\log 8}} \cr
& = \frac{1}{2} \cr} $$

Answer: Option A. -> a + b = 1
$$\eqalign{
& \log {a \over b} + \log {b \over a} = \log \left( {a + b} \right) \cr
& \Rightarrow \log \left( {a + b} \right) = \log \left( {{a \over b} \times {b \over a}} \right) = \log 1 \cr
& So,a + b = 1 \cr} $$

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

Answer: Option A. -> p2 + q2
$$\eqalign{
& {\text{Given}}, \cr
& {\log _{10}}a = p,\,{\log _{10}}b = q \cr
& {\log _{10}}\left( {{a^p}{b^q}} \right) = {\log _{10}}{a^p} + {\log _{10}}{b^q} \cr
& = p{\log _{10}}a + q{\log _{10}}b \cr
& = {p^2} + {q^2} \cr} $$

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