Question

Find the value of \[lo{g_3}\left( 5 \right){\text{ }}lo{g_{25}}\left( {27} \right).\]

Answer 1

Hint: We have to separate the base and radicand and then simplify using logarithmic formula. We will convert the base of log so that each log can have the same base. Some formulae of logarithm are given below:
\[{\log _a}b = \dfrac{{{{\log }_b}}}{{{{\log }_a}}}\], where is called radicand and is called base.
\[\log {(a)^m} = m\,\,\log a\]

Complete step by step answer:
Given: \[{\log _3}5 \times {\log _{25}}27\]
\[\left( 1 \right)\] As both logarithmic terms are having different bases. So, first separate base and radicand by using formula,
\[{\log _a}b = \dfrac{{\log \,b}}{{\log \,a}}\]
\[\left( 2 \right)\] \[{\log _3}5 \times {\log _{25}}27\]
\[{\log _3}5 \times {\log _{25}}27 = \dfrac{{\log \,\,5}}{{\log \,\,3}} \times \dfrac{{\log \,\,27}}{{\log \,\,25}}\] \[\left( {\because \,{{\log }_a}b = \dfrac{{\log b}}{{\log a}}} \right)\]
We will find prime factors of number$27 = 3 \times 3 \times 3 = {3^3}$and \[25 = 5 \times 5 = {5^2}\]
\[ = \dfrac{{\log \,\,5}}{{\log \,\,3}} \times \dfrac{{\log {{(3)}^3}}}{{\log {{(5)}^2}}}\]
\[ = \dfrac{{\log \,\,5}}{{\log \,\,3}} \times \dfrac{{3\log \,(3)}}{{2\log \,\,5}}\] \[\left( {\because \log {a^m} = m\,\,\log a} \right)\]
\[ = \dfrac{3}{2}\]
Hence, \[lo{g_3}5 \times lo{g_{25}}27 = \dfrac{3}{2}\]

Additional Information: Logarithmic functions are the inverse of exponential functions. The inverse of the exponential function\[y = {a^x}\,is{\text{ }}x = {a^y}\]. The logarithmic function \[y = lo{g_a}x\] is said to be equivalent to the exponential equation \[x = {a^y},y = lo{g_a}x\] only under the following conditions:\[x = {a^y},a > 0\], and\[a\; \ne 1\] . It is called the logarithmic function with base $a$.

Note: Students should take care about the base of log. In logarithmic terms, simplification is only possible when terms have the same base. I