Question

How do you find the value of ${{\log }_{7}}\left( 42 \right)$

Answer 1

Hint: To evaluate ${{\log }_{7}}\left( 42 \right)$ we will use the properties of logarithm. Now we know that ${{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}$ . Also we know that $\log \left( a\times b \right)=\log \left( a \right)+\log \left( b \right)$ and log $\log {{\left( a \right)}^{m}}=m\times \log \left( a \right)$ . We will use this properties to simplify the given expression.
Now we have $\log \left( 2 \right)=0.3$ and $\log \left( 7 \right)=0.84$. We will use this value and find the value of the given expression

Complete step-by-step answer:
Let us first understand Logarithm.
Let us say we have an expression ${{b}^{y}}=x.$
Then in terms of Logarithm we have ${{\log }_{b}}x=y$
Here b is called base of Logarithm.
10 is the standard base of logarithm and log with base 10 can be written as just log.
Hence ${{\log }_{10}}100$ is written as $\log 100$ .
Now consider the given expression ${{\log }_{7}}\left( 42 \right)$
Now since the base of logarithm is 7 we will change the base to 10 which is the standard base of log, with the help of property ${{\log }_{a}}b=\dfrac{{{\log }_{c}}b}{{{\log }_{c}}a}$ .
Hence, using this we get ${{\log }_{7}}42=\dfrac{\log 42}{\log 7}$
Now we know that $\log \left( 42 \right)=\log \left( 7\times 6 \right)$
Hence using this we get,
${{\log }_{7}}42=\dfrac{\log \left( 7\times 6 \right)}{\log 7}$
Now we know that $\log \left( a\times b \right)=\log \left( a \right)+\log \left( b \right)$ . Using this we get,
$\begin{align}
  & {{\log }_{7}}\left( 42 \right)=\dfrac{\log \left( 7 \right)+\log \left( 6 \right)}{\log \left( 7 \right)} \\
 & \Rightarrow {{\log }_{7}}\left( 42 \right)=\dfrac{\log \left( 7 \right)}{\log \left( 7 \right)}+\dfrac{\log \left( 6 \right)}{\log \left( 7 \right)} \\
\end{align}$
Now writing 6 as $2\times 3$ we get.
${{\log }_{7}}\left( 42 \right)=1+\dfrac{\log \left( 2\times 3 \right)}{\log \left( 7 \right)}$
Now again we know that $\log \left( a\times b \right)=\log \left( a \right)+\log \left( b \right)$ Hence, using this property we get
${{\log }_{7}}\left( 42 \right)=1+\dfrac{\log 2+\log 3}{\log 7}$
Now we know that the value of $\log 2=0.3$ , $\log 3=0.477$ and the value of $\log 7=0.84$
Hence Substituting this in ${{\log }_{7}}\left( 42 \right)=1+\dfrac{\log 2+\log 3}{\log 7}$ we get,
\[\begin{align}
  & {{\log }_{7}}\left( 42 \right)=1+\dfrac{0.477+0.3}{0.84} \\
 & \Rightarrow {{\log }_{7}}\left( 42 \right)=1+\dfrac{0.777}{0.84} \\
 & \Rightarrow {{\log }_{7}}\left( 42 \right)=1+0.925 \\
 & \therefore {{\log }_{7}}\left( 42 \right)=1.925 \\
\end{align}\]
Hence the value of \[{{\log }_{7}}\left( 42 \right)\] is 1.925.

Note: Now in this problem we can directly write $\log \left( 42 \right)=\log \left( 7\times 2\times 3 \right)$ and then solve for
$\dfrac{\log 7+\log 3+\log 2}{\log 7}=\dfrac{0.3+0.477+0.84}{0.84}$ . Also note that log is logarithm with base 10 and ln is logarithm with base e where e is nothing but 2.71828.