Inequality with a mixture of logs and trig Please help me solve this (rather difficult) inequality (or give a hint as to how to do it) -- thanks!
$$
    \frac{\log _{3-x}(4 x-5) \cdot \log _{4 x}\left(\log _2(7)-x\right)}{\cos  x}\leq 0 
$$
 A: Hint
I would suggest you go back to logarithms in base $e$. So, your expression $$\frac{\log _{3-x}(4 x-5) \cdot \log _{4 x}\left(\log _2(7)-x\right)}{\cos  x}=\frac{\log (4 x-5) \log \left(\frac{\log (7)}{\log (2)}-x\right) \sec (x)}{\log
   (3-x) \log (4 x)}$$ So, you need now to look at the sign of each piece in the proper intervals.
I am sure that you can take from here.
A: Firstly, $3-x$, $4x$,$4x-5$ and $\log_27-x$ must be positive.
Secondly, we want the sign of the product of five numbers.
$$\frac{\log(4x-5)\log(\log_2(7)-x)}{(\cos x)\log(3-x)\log(4x)}$$
It is simple to find where each of the factors changes sign.  Put out a numberline, and see where the result is positive and negative.
A: Hopefully Claude's and Michael's answers are enough for you to get the answer.  If you get stuck, here are a few more hints:


*

*You need to consider more than just zeros and points of discontinuity for each factor in the expression on the left... keep in mind that the domain of the expression is restricted by its factors.  Also, be careful whether the sets used to present your solution are open or closed.  

*Given the nature of this problem, I'm going to assume you'll want to be able to find your answer without a calculator.  In this case, you will find that you are going to have to determine which of $\pi/2$ or $\log_{2}7-1$ is greater (they both lie between $3/2$ and $2$).  Assuming you don't know $\log_{2}7$  by memory, use various fractional powers of $2$ to help you find bounds for the latter quantity.
