Question

Solve the following equation: ${\log _{0.1}}\sin 2x + \log \cos x - \log 7 = 0$.

Answer 1

Hint: In this question, we are given an equation in terms of log and we have been asked to solve it and find the value of $x$. Start by simplifying the first term of the equation and make sure that the base of all the terms in the equation are the same. Then, simplify using the properties of log. Use $\log 1 = 0$ to compare both the sides. After the equation has been simplified, raise inverse trigonometry to find the answer.

Complete step-by-step solution:
We have been given an equation in terms of log. We are required to solve it and find the value of $x$.
$ \Rightarrow {\log _{0.1}}\sin 2x + \log \cos x - \log 7 = 0$
We will at first simplify the first term of the given equation -${\log _{0.1}}\sin 2x$.
We must know that when no base is given, it is assumed that the base is $10$. So, we will simplify the first term of the given equation such that its base is also $10$.
$ \Rightarrow {\log _{0.1}}\sin 2x$
We will first use base conversion formula - ${\log _a}b = \dfrac{{{{\log }_c}b}}{{{{\log }_c}a}}$.
$ \Rightarrow {\log _{0.1}}\sin 2x$$ = \dfrac{{\log \sin 2x}}{{\log 0.1}}$ …. (1)
We can write $\log 0.1$ as $\log \dfrac{1}{{10}}$. Using the formula, $\log \dfrac{m}{n} = \log m - \log n$ to write $\log \dfrac{1}{{10}}$ -
$ \Rightarrow \log \dfrac{1}{{10}} = \log 1 - \log 10$
$ \Rightarrow \log \dfrac{1}{{10}} = 0 - \log 10$ …. $\left( {\log 1 = 0} \right)$
$ \Rightarrow \log \dfrac{1}{{10}} = - \log 10$ …. (2)
Putting equation (2) in (1),
$ \Rightarrow {\log _{0.1}}\sin 2x$$ = \dfrac{{\log \sin 2x}}{{ - \log 10}}$
Now we will again use the base conversion formula - $\dfrac{{{{\log }_c}b}}{{{{\log }_c}a}} = {\log _a}b$.
$ \Rightarrow {\log _{0.1}}\sin 2x = - {\log _{10}}\sin 2x$
Now we can write $ - {\log _{10}}\sin 2x$ as $ - \log \sin 2x$. Our equation has now become –
$ \Rightarrow - \log \sin 2x + \log \cos x - \log 7 = 0$
Now we will use properties $\log m - \log n = \log \dfrac{m}{n}$ and $\log m + \log n = \log \left( {mn} \right)$ to simplify the equation that we have.
$ \Rightarrow \log \cos x - \left( {\log \sin 2x + \log 7} \right) = 0$
Using $\log m + \log n = \log \left( {mn} \right)$,
$ \Rightarrow \log \cos x - \log 7\sin 2x = 0$
Now, we will use $\log m - \log n = \log \dfrac{m}{n}$ -
$ \Rightarrow \log \dfrac{{\cos x}}{{7\sin 2x}} = 0$
We know that $\log 1 = 0$. So, we will substitute this on RHS.
$ \Rightarrow \log \dfrac{{\cos x}}{{7\sin 2x}} = \log 1$
Comparing both the sides,
$ \Rightarrow \dfrac{{\cos x}}{{7\sin 2x}} = 1$
We can also write $\sin 2x = 2\sin x\cos x$.
$ \Rightarrow \dfrac{{\cos x}}{{14\sin x\cos x}} = 1$
On simplifying we will get,
$ \Rightarrow \dfrac{1}{{14\sin x}} = 1$
$ \Rightarrow \dfrac{1}{{14}} = \sin x$
Now, in order to find the value of $x$, we will raise inverse trigonometry.
$ \Rightarrow x = {\sin ^{ - 1}}\dfrac{1}{{14}}$

The simplified form of the given equation is $x = {\sin ^{ - 1}}\dfrac{1}{{14}}$.

Note: It must be noted that we could not directly use the properties of log in the equation given in the question as the base of the terms were not the same. So, if this case arises, our first step will be to make the base of all the terms the same. If the base is not given, always assume the base to be $10$. If it is given, make sure that all the terms have that base.