Question

How do you condense $In\,3 - 2\,In\,5 + In\,10?$

Answer 1

In order to condense the given expression we will be using some basic rules of logarithms that are nothing but rules of addition, rule of subtraction, rule of multiplication, rule of division and rule of power. After applying this rule we will simplify it and hence, we will get our required answer.

Complete Step by Step Solution:
In this question we have given. $In3-2In5+In10$ ,........(1)
Since we know that if we have $a In(b)$ then it can be written as $In{{b}^{a}}$
So using this we can rewrite expression $(1)$ as
$\Rightarrow \ln 3-\ln {{5}^{2}}+\ln 10\text{ }\left( \because a\ln b=\ln {{b}^{2}} \right)$ ……(2)
Since we know that if we have $In\left( a-b \right)$ then it can be written as $In\left( \dfrac{a}{b} \right)$.
So using this we can rewrite expression $\left( 2 \right)$ as
$\Rightarrow \ln \left( \dfrac{3}{{{5}^{2}}} \right)+\ln 10\text{ }\left( \because \ln \left( a-b \right)=\ln \left( \dfrac{a}{b} \right) \right)$ .............(3)
Also, we know that
If we have $In\left( a+b \right)$ then it can be written as $In\left( ab \right)$.
So, using this we can rewrite expression $\left( 3 \right)$ as
$\Rightarrow \ln \left( \dfrac{3}{{{5}^{2}}}\times 10 \right)\text{ }\left( \because \ln \left( a+b \right)=\ln \left( ab \right) \right)$ .............(4)
On further simplifying expression $\left( 4 \right)$ we get,
$\Rightarrow \ln 3-2\ln 5+\ln 10=\ln \left( \dfrac{6}{5} \right)$
Hence, $In\,3-2\,In5+In10$ can be written as $In\left(\dfrac{6}{5} \right)$ which is our required answer.

Note: There are some rules of logarithms to always remember which are as follows!
1. That is $\log \left( xy \right)=\log \left( x \right)+\log \left( y \right)$
2. logs of the same base can be subtract by dividing the arguments that is $\log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right)$
Some important values of $\log s$.
1. ${{\log }_{b}}\left( 1 \right)\,\,=0$
2. ${{\log }_{b}}\left( b \right)\,\,=1$
If we want to change the base of given $\log$ then the rule is as follows:
${{\log }_{b}}\left( M \right)=\dfrac{{{\log }_{a}}\left( M \right)}{{{\log }_{a}}\left( b \right)}$