Question

How do you evaluate $\log 300-\log 3$?

Answer 1

Hint: Now to solve the given equation we will first use the division rule which states $\log \left( \dfrac{x}{y} \right)=\log x-\log y$ to simplify the equation. Now we will solve this equation by using the definition of log function.

Complete step by step solution:
Now first let us understand the concept of logarithms. Now in general the logarithm is written as ${{\log }_{b}}x$ . What the function gives us is the index to which b must be raised so that we get x.
Let us understand this with an example. Now let us consider the equation ${{10}^{2}}=100$ . Now this can be written in logarithmic form as ${{\log }_{10}}100=2$ .
Now note that if the base of logarithm is not mentioned then the base is taken as 10.
Now let us understand a few properties of logarithm.
The index property of logarithm states that $\log {{x}^{n}}=n\log x$
Similarly by multiplication property of logarithm we have $\log \left( x\times y \right)=\log x+\log y$ and by division property we have $\log \left( \dfrac{x}{y} \right)=\log x-\log y$
Now consider the given equation $\log 300-\log 3$ .
Now using the division rule we know that $\log \left( \dfrac{x}{y} \right)=\log x-\log y$. Hence using this we get,
$\begin{align}
  & \Rightarrow \log 300-\log 3=\log \left( \dfrac{300}{3} \right) \\
 & \Rightarrow \log 300-\log 3=\log 100 \\
\end{align}$
Now we know that the base of logarithm is 10.
Now let $\log 100=n$ . Then writing this in index form we get,
$\begin{align}
  & \Rightarrow {{10}^{n}}=100 \\
 & \Rightarrow n=2 \\
\end{align}$
Hence the solution of the given function is n = 2.

Note: Now note that in general if the base is not given then we take the base of logarithm as 10. If the base is taken as the constant e then the logarithm is called natural logarithm and written as ln. Also note that ${{\log }_{b}}0$ is not defined.