Question

Calculate \[{\log _9}40\] if \[\log 15 = a\] and \[{\log _{20}}50 = b\]

Answer 1

Hint: Here we will use the basic properties of the logarithmic operations to solve the \[{\log _9}40\]. So firstly we will find the value of \[\log 2\& \log 3\] in terms of \[a\& b\] by solving the because by simplifying the \[{\log _9}40\] term we will get the final equation in terms of \[\log 2\& \log 3\].

Complete step-by-step answer:
It is given that \[\log 15 = a\].
Now we will find the value of \[\log 3\] by solving the given equation \[\log 15 = a\]. Therefore by using the basic properties of the log function we get
\[\log 15 = \log \left( {\dfrac{{3 \times 10}}{2}} \right) = \log \left( {3 \times 10} \right) - \log 2 = \log 3 + \log 10 - \log 2 = a\]
We know that value of \[\log 10\] is equals to 1. Therefore by solving the above equation, we get
\[ \Rightarrow \log 15 = \log 3 + 1 - \log 2 = a\]
\[ \Rightarrow \log 3 = a + \log 2 - 1\]…………………..\[\left( 1 \right)\]
It is given that \[{\log _{20}}50 = b\].
Now we will find the value of \[\log 2\] by solving the given equation \[{\log _{20}}50 = b\]. Therefore by using the basic properties of the log function we get
\[{\log _{20}}50 = \dfrac{{\log 50}}{{\log 20}} = \dfrac{{\log \left( {\dfrac{{100}}{2}} \right)}}{{\log \left( {2 \times 10} \right)}} = \dfrac{{\log 100 - \log 2}}{{\log 2 + \log 10}} = \dfrac{{\log {{10}^2} - \log 2}}{{\log 2 + \log 10}} = \dfrac{{2\log 10 - \log 2}}{{\log 2 + \log 10}} = b\]
\[ \Rightarrow \dfrac{{2\log 10 - \log 2}}{{\log 2 + \log 10}} = b\]
We know that the value of \[\log 10\] is equal to 1. Therefore by solving the above equation, we get
\[ \Rightarrow \dfrac{{2 \times 1 - \log 2}}{{\log 2 + 1}} = \dfrac{{2 - \log 2}}{{\log 2 + 1}} = b\]
\[ \Rightarrow 2 - \log 2 = b\left( {\log 2 + 1} \right) = b\log 2 + b\]
By solving this we will get the value of \[\log 2\], we get
\[ \Rightarrow b\log 2 + \log 2 = 2 - b\]
\[ \Rightarrow \log 2 \times \left( {1 + b} \right) = 2 - b\]
\[ \Rightarrow \log 2 = \dfrac{{2 - b}}{{1 + b}}\]…………………..\[\left( 2 \right)\]
Now we will find the value of \[{\log _9}40\]. Therefore, we will simplify the main equation i.e. \[{\log _9}40\], we get
\[ \Rightarrow {\log _9}40 = \dfrac{{\log 40}}{{\log 9}} = \dfrac{{\log \left( {{2^2} \times 10} \right)}}{{\log {3^2}}} = \dfrac{{\log {2^2} + \log 10}}{{2\log 3}} = \dfrac{{2\log 2 + 1}}{{2\log 3}}\]
Now we will put the value of \[\log 2\& \log 3\] from the equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\]. Therefore, we get
\[ \Rightarrow {\log _9}40 = \dfrac{{2\log 2 + 1}}{{2\log 3}} = \dfrac{{2\left( {\dfrac{{2 - b}}{{1 + b}}} \right) + 1}}{{2\left( {a + \log 2 - 1} \right)}} = \dfrac{{2\left( {\dfrac{{2 - b}}{{1 + b}}} \right) + 1}}{{2\left( {a + \left( {\dfrac{{2 - b}}{{1 + b}}} \right) - 1} \right)}} = \dfrac{{\dfrac{{4 - 2b}}{{1 + b}} + 1}}{{2a + \dfrac{{4 - 2b}}{{1 + b}} - 2}} = \dfrac{{\dfrac{{4 - 2b + \left( {1 + b} \right)}}{{1 + b}}}}{{\dfrac{{2a\left( {1 + b} \right) + 4 - 2b - 2\left( {1 + b} \right)}}{{1 + b}}}}\]
Now we will simplify the above equation, we get
\[ \Rightarrow {\log _9}40 = \dfrac{{\dfrac{{4 - 2b + \left( {1 + b} \right)}}{{1 + b}}}}{{\dfrac{{2a\left( {1 + b} \right) + 4 - 2b - 2\left( {1 + b} \right)}}{{1 + b}}}} = \dfrac{{4 - 2b + \left( {1 + b} \right)}}{{2a\left( {1 + b} \right) + 4 - 2b - 2\left( {1 + b} \right)}} = \dfrac{{5 - b}}{{2a + 2ab + 4 - 2b - 2 - 2b}} = \dfrac{{5 - b}}{{2a + 2ab + 2 - 4b}}\]
Hence, \[{\log _9}40\] is equal to \[\dfrac{{5 - b}}{{2a + 2ab + 2 - 4b}}\].

Note: We should know that the value inside the log function should never be zero or negative it should always be greater than zero. We should know that the value of the \[\log 10\] is equal to 1. We should be simplifying the equation carefully and apply the properties of the log function accurately. We should also know the basic properties of the log functions.
\[\begin{array}{l}
\log a + \log b = \log ab\\
\log a - \log b = \log \dfrac{a}{b}\\
{\log _a}b = \dfrac{{\log b}}{{\log a}}
\end{array}\]