# Tricky Logarithmic Inequality Problem

I am having a problem solving this question - If $\log_{\frac{1}{\sqrt{2}} }{\sin{x}}>0$,

$x\in [0,4\pi]$,then number of values for chating which are integral multiples of $\pi/4$,is

A-6

B-12

C-3

D-none of these

My book says that $x$ belongs to $(0,\pi)\cup(2\pi,3\pi)....$ How???

Then if $x$ does belong to that then we get 6 solutions which is the answer using inequality $\sin{x}<1$.

BUT how does $x$ belong to that inequality... Please help...

Let $\displaystyle\log_{\frac1{\sqrt2}}(\sin x)=y \implies\sin x=\left(\frac1{\sqrt2}\right)^y=2^{-y/2}$

Now $0<y<\infty\iff0>-y/2>-\infty\iff1>2^{-y/2}>0\implies1>\sin x>0$

Using All-Sin-Tan-Cos Rule, $x$ must lie in the first and second Quadrant i.e., $2n\pi<x<(2n+1)\pi$ where $n$ is any integer

• It should be 1st or 2nd quadrant – Jai Mahajan Mar 22 '15 at 12:20
• @user166748, Please find rectified version – lab bhattacharjee Mar 22 '15 at 12:21
• Thanks a lot... Now the answer is perfect :) – Jai Mahajan Mar 22 '15 at 12:26

In answer the the "How???" question, the logarithm of a number is defined only if the number is positive, and $\sin x\ge0$ only on $(0,\pi)$ and $(2\pi,3\pi)$ in the interval $0\le x\le4\pi$. So you do get $6$ candidates for the multiples of $\pi/4$. But for two of them, namely $x=2\pi/4$ and $x=10\pi/4$, $\sin x=1$, and the logarithm of $1$ is $0$ in any base. That leaves only $4$ candidates.

At this point there are two ways to complete the problem. You can either note that $\sin x=1/\sqrt2$ for all four candidates, so that $\log_{1/\sqrt2}\sin x=1\gt0$ all four times, so that answer D, "none of these," is the correct choice. Or you can simply note that you've already eliminated answers A and B, while $\sin(x+2\pi)=\sin x$ means that the values you're considering come in equal pairs, so the odd number $3$ is not a possible answer, thus eliminating C.