# Use the properties of the natural logarithms to simplify the

Use the properties of the natural logarithms to simplify the following expressions:
$$\displaystyle{a}{)}{\ln{{\sin{\theta}}}}-{\ln{{\frac{{{\sin{\theta}}}}{{{5}}}}}}$$

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Ceitheart
Step1
Logarithm is inverse of exponentiations. A logarithm is easy method to express large number and to perform arithmetic operation on them. Multiplication and division can be written in form of addition and subtraction while operating logarithms.
There are various rules involved in performing logarithmic operations, some required for question are as follows
$$\displaystyle\text{Product rule }\ \rightarrow{{\log}_{{a}}{m}}+{{\log}_{{a}}{n}}={{\log}_{{a}}{m}}{n}$$
$$\displaystyle\text{Product rule }\ \rightarrow{{\log}_{{a}}{m}}-{{\log}_{{a}}{n}}={{\log}_{{a}}{\left({\frac{{{m}}}{{{n}}}}\right)}}$$
Step 2
The given expression is $$\displaystyle{\ln{{\left({\sin{\theta}}\right)}}}-{\ln{{\left({\frac{{{\sin{\theta}}}}{{{5}}}}\right)}}}$$ , use division rule to simplify the expression
$$\displaystyle{\ln{{\left({\sin{\theta}}\right)}}}-{\ln{{\left({\frac{{{\sin{\theta}}}}{{{5}}}}\right)}}}={\ln{{\left({\frac{{{\sin{\theta}}}}{{{\frac{{{\sin{\theta}}}}{{{5}}}}}}}\right)}}}={\ln{{5}}}$$
Therefore, simplified version $$\displaystyle{\ln{{\left({\sin{\theta}}\right)}}}-{\ln{{\left({\frac{{{\sin{\theta}}}}{{{5}}}}\right)}}}\ \text{ is equal to }\ {\ln{{5}}}$$
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Hiroko Cabezas
$$\displaystyle{\ln{{\sin{\theta}}}}-{\ln{{\frac{{{\sin{\theta}}}}{{{5}}}}}}={\ln{{\sin{\theta}}}}-{\left({\ln{{\sin{\theta}}}}-{\ln{{5}}}\right)}={\ln{{5}}}$$