Question

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

Answer 1

Hint:
Here we will use logarithmic rules log have a lot of rules but we use some related rules for logarithmic. Base values and log values given here we substitute the $a$ and $b$ values we will get the answer.

Formula used:
$\log \dfrac{a}{b} = \log a - \log b$
$\log (a \times b) = \log a + \log b$

Complete step by step solution:
Given in question is
$b = {\log _{20}}50$
This equation is change into rule
$ \Rightarrow \dfrac{{\log 50}}{{\log 20}}$ (Here $50$ is change to $\dfrac{{100}}{2}$ and $20$ is change to $(2 \times 10)$)
Using some rule above equation is change to
$ \Rightarrow \dfrac{{\log \dfrac{{100}}{2}}}{{\log (2 \times 10)}}$$ = \dfrac{{\log 100 - \log 2}}{{\log 2 + \log 10}}$ (Here $\log \dfrac{a}{b} = \log a - \log b$ and $\log (a \times b) = \log a + \log b$ using this formula)
$ \Rightarrow \dfrac{{\log {{10}^2} - \log 2}}{{\log 2 + \log 10}}$ (Here $100$ will be change ${10^2}$ )
$ \Rightarrow \dfrac{{2\log 10 - \log 2}}{{\log 2 + \log 10}}$ (Here we are taking $\log 10$ as ${\log _{10}}10$ so ${\log _{10}}10 = 1$)
$b = \dfrac{{2 - \log 2}}{{1 + \log 2}}$
Here $1 + \log 2$ is going to left hand side
$b + b\log 2 = 2 - \log 2$
$\log 2 = \dfrac{{2 - b}}{{b + 1}}$
Next, we will take $a$ value
$a = \log 15$
Here $\log 15$is change to $\log 3 + \log 5$
$ = \log 3 + \log 10 - \log 2$ (here $\log 5$ is change to $\log \dfrac{{10}}{2}$ then this will be change to $\log 10 - \log 2$)
$ = \log 3 + 1 - \log 2$ (Here $\log 10$ is $1$ )
$ = \log 3 = a + \log 2 - 1$
Now we take
${\log _9}40 = \dfrac{{\log 40}}{{\log 9}}$
$ = \dfrac{{\log (10 \times {2^2})}}{{\log {3^2}}}$ ($\log 40$ is change to $\log (10 \times {2^2})$)
$ = \dfrac{{\log 10 + 2\log 2}}{{2\log 3}}$
$ = \dfrac{{1 + 2\log 2}}{{2\log 3}}$
Substitute the $\log 2$ value and $\log 3$ value in above equation
$ = \dfrac{{1 + 2(2 - b)}}{{a + \dfrac{{2 - b}}{{b + 1}} - 1}}$
Simplify the above equation we will get the answer for the question
$ = \dfrac{{5 - b}}{{ab + a + 1 - 2b}}$

So, the answer for the question is $ = \dfrac{{5 - b}}{{ab + a + 1 - 2b}}$

Additional Information:
Logarithms and exponentials with the same base cancel each other. This is true because logarithms and exponentials are inverse operations much like the same way multiplication and division are inverse operations, and addition and subtraction are inverse operations.

Note:
Here we will be using many formulas on these questions. Now we have many formulas in case multiple values we will plus the values incase divide the values we will be minus the value. Here, assuming we have a lot of value. While solving this type of question we will use many formulas.