Question

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

Answer 1

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.

\log a + \log b = \log ab\\\ \log a - \log b = \log \dfrac{a}{b}\\\ {\log _a}b = \dfrac{{\log b}}{{\log a}} \end{array}$$