Simplify the following logarithmic expressions.
\log_3\left(2\right) + \log_3\left(6\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(a\right)+ \log_c\left(b\right)=\log_c\left(ab\right)
Therefore:
\log_3\left(2\right) + \log_3\left(6\right) = \log_3\left(2\times 6\right)
\log_3\left(2\right) + \log_3\left(6\right)=\log_3\left(12\right)
\log_2\left(14\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(ab\right) = \log_c\left(a\right)+ \log_c\left(b\right)
Therefore:
\log_2\left(14\right)= \log_2\left(2 \times 7\right)
\log_2\left(14\right)=\log_2\left(2\right) + \log_2\left(7\right)
Notice that:
\log_2\left(2\right)=1
Therefore:
\log_2\left(14\right)=1+\log_2\left(7\right)
\log_{18}\left(3\right) + \log_{18}\left(6\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(a\right)+ \log_c\left(b\right) =\log_c\left(ab\right)
Therefore:
\log_{18}\left(3\right)+ \log_{18}\left(6\right)=\log_{18}\left(3\times 6\right)
\log_{18}\left(3\right)+ \log_{18}\left(6\right)= \log_{18}\left(18\right)
\log_{18}\left(3\right) + \log_{18}\left(6\right) =1
\log_3\left(27\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(ab\right) = \log_c\left(a\right)+ \log_c\left(b\right)
Therefore:
\log_3\left(27\right)= \log_3\left(9 \times 3\right)
\log_3\left(27\right)=\log_3\left(9\right) + \log_3\left(3\right)
Also:
\log_3\left(9\right)=\log_3\left(3 \times 3\right)
\log_3\left(9\right) = \log_3\left(3\right) + \log_3\left(3\right)
Therefore:
\log_3\left(27\right) = \log_3\left(3\right) + \log_3\left(3\right)+ \log_3\left(3\right)
Notice that:
\log_3\left(3\right)=1
Therefore:
\log_3\left(27\right)=3
\log_3\left(48\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(ab\right) = \log_c\left(a\right)+ \log_c\left(b\right)
Therefore:
\log_3\left(48\right)= \log_3\left(3 \times 16\right)
\log_3\left(48\right)= \log_3\left(3\right) + \log_3\left(16\right)
Notice that:
\log_3\left(3\right)=1
Therefore:
\log_3\left(48\right)= 1+ \log_3\left(16\right)
\log_9\left(5\right) + \log_9\left(8\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(a\right)+ \log_c\left(b\right) =\log_c\left(ab\right)
Therefore:
\log_{9}\left(5\right)+ \log_{9}\left(8\right)=\log_{9}\left(5\times 8\right)
\log_{9}\left(5\right)+ \log_{9}\left(8\right)= \log_{9}\left(40\right)
\log_5\left(36\right)
For positive real numbers a,b and c, where c \ne1 :
\log_c\left(ab\right) = \log_c\left(a\right)+ \log_c\left(b\right)
Therefore:
\log_5\left(36\right)= \log_5\left(6\times 6\right)
\log_5\left(36\right)= \log_5\left(6\right) + \log_5\left(6\right)
\log_5\left(36\right)= 2\log_5\left(6\right)