Question

Which of the following is/are positive?
A. ${\log _{\sin 1}}\tan 1$
B. ${\log _{\cos 1}}\left( {1 + \tan 3} \right)$
C. ${\log _{{{\log }_{10}}5}}\left( {\cos \theta + \sec \theta } \right)$
D. ${\log _{\tan {{15}^o}}}\left( {2\sin {{18}^o}} \right)$

Answer 1

Hint: In this question, from the given functions, we have to tell the positive ones. We have that, in ${\log _a}x$ , if $a > 1$ and $x > 1$ or $a < 1$ and $x < 1$ then, ${\log _a}x > 0$ , otherwise, it is negative.
$\tan \theta $ is always an increasing function, whereas, $\sin \theta $ and $\cos \theta $ are neither increasing nor decreasing.

Complete answer:
Given are four logarithmic functions.
To tell which of these are positive.
Firstly, we know, the graph of ${\log _a}x$ , when $a > 1$ is given by

Then, ${\log _a}x > 0$ , when $x > 1$ .
And when $a < 1$ , the graph is given by

Then, ${\log _a}x > 0$ , when $x < 1$ .
Now, consider the function, ${\log _{\sin 1}}\tan 1$ , we have, $\dfrac{\pi }{2} = \dfrac{{3.14}}{2} > 1$ , so, $\sin \dfrac{\pi }{2} > \sin 1$ which gives $\sin 1 < 1$ .
And, $\dfrac{\pi }{4} = \dfrac{{3.14}}{4} < 1$ , so, $\tan \dfrac{\pi }{4} < \tan 1$ which give $\tan 1 > 1$ . Now, $\sin 1 < 1$ and $\tan 1 > 1$ , hence, ${\log _{\sin 1}}\tan 1 < 0$ .
Now, consider the function, ${\log _{\cos 1}}\left( {1 + \tan 3} \right)$ , $0 < 1 < \dfrac{\pi }{2}$ , so, $\cos 0 > \cos 1 > \cos \dfrac{\pi }{2}$ , since, the function is decreasing for $0 < \theta < \pi $ , hence, $\cos 1 < 1$
And, $\dfrac{\pi }{2} < 3 < \pi $ , which means it is in the second quadrant, and the second quadrant $\tan $ is negative, which means $1 + \tan 3 < 1$ .
Now, $\cos 1 < 1$ and $1 + \tan 3 < 1$ , hence, ${\log _{\cos 1}}\left( {1 + \tan 3} \right) > 0$ .
Now, consider the function, ${\log _{\tan {{15}^o}}}\left( {2\sin {{18}^o}} \right)$ , we know, $\sin {18^o} < \sin {30^o}$ , which means, $\sin {18^o} < \dfrac{1}{2}$ i.e., $2\sin {18^o} < 1$ .
And, we know, $\tan {15^o} < \tan {45^o}$ , which gives, $\tan {15^o} < 1$ . Now, $2\sin {18^o} < 1$ and $\tan {15^o} < 1$ , so, ${\log _{\tan {{15}^o}}}\left( {2\sin {{18}^o}} \right) > 0$ .
Now, at last, consider the function, ${\log _{{{\log }_{10}}5}}\left( {\cos \theta + \sec \theta } \right)$ , we know, ${\log _{10}}5 < {\log _{10}}10$ i.e., ${\log _{10}}5 < 1$ .
And, we know, the Arithmetic mean is always greater than the geometric mean, therefore, $\dfrac{{\left( {\cos \theta + \sec \theta } \right)}}{2} > {\left( {\cos \theta \cdot \sec \theta } \right)^{\dfrac{1}{2}}}$ , which gives, $\cos \theta + \sec \theta > 2 > 1$ .
Now, ${\log _{10}}5 < 1$ and $\cos \theta + \sec \theta > 2 > 1$ , hence, ${\log _{{{\log }_{10}}5}}\left( {\cos \theta + \sec \theta } \right) < 0$ .
Thus, ${\log _{\tan {{15}^o}}}\left( {2\sin {{18}^o}} \right)$ and ${\log _{\cos 1}}\left( {1 + \tan 3} \right)$ are positive.

Therefore, the correct option is B and C

Note: We know that, $\sec \theta $ is reciprocal of $\cos \theta $ .
“Arithmetic mean is always greater than the geometric mean” is true for every function or equation.
For checking, if a logarithmic function is positive or negative, we have to check if $a$ and $x$ are greater than or less than $1$ , if the sign of $a$ and $x$ are same i.e., either both greater than $1$ or both less than $1$, then, the logarithmic function is positive, and if the sign of $a$ and $x$ are not same, i.e., one is greater than $1$ and other is less than $1$, then the logarithmic function is negative.