How to solve for $\frac{3x}{16}\left(\frac{\log 2}{\log x}\right) > 1$ I took the equation as:
$$3x > 16\left(\frac{\log x}{\log 2}\right)$$
Trying out different values, I found that it is true for $x=32$ since:
$$96 > 16\left(\frac{\log 32}{\log 2}\right) = 80 $$
To check if the function is increasing, I take the derivative:
$$\frac{d}{dx}\left(\frac{3x\log 2}{16 \log x}\right) = \frac{3\log 2(\log x - 1)}{16 \log^2 x}$$
Now, I am stuck.  I have two questions:


*

*How does one determine the first $x$ where $3x > 16\left(\frac{\log x}{\log 2}\right)$?  I couldn't figure out so I just tried out numbers.

*For such a complex derivative, how does one determine if it increasing for $x > 32$ or if it increases only for an interval such as $[32, 32+k]$?


When I plug in $32$ to the derivative, for example, I am calculating:
$$\frac{3 \log 2(\log 32 - 1)}{16\log^2(32)} > 0.0266$$
so, the function is increasing at $x =32$.  Am I right?
 A: Almost as you did, considering the function $$f(x)=\frac{3x}{16}\left(\frac{\log 2}{\log x}\right)-1$$ Computing the derivatives $$f'(x)=\frac{3 \log (2) (\log (x)-1)}{16 \log ^2(x)}$$ $$f''(x)=-\frac{3 \log (2) (\log (x)-2)}{16 x \log ^3(x)}$$ the first derivative cancels when $x=e$. At this point $$f(e)=\frac{3}{16} e \log (2)-1\approx -0.646718$$ $$f''(e)=\frac{3 \log (2)}{16 e} >0$$ which means that this is a minimum.
So, the equation $f(x)=0$ has two roots which, as  Donald Splutterwit answered, express in terms of Lambert function. $$x_1=-\frac{16}{3 \log (2)} W\left(-\frac{3 \log (2)}{16}\right)$$ $$x_2=-\frac{16 }{3 \log (2)}W_{-1}\left(-\frac{3 \log (2)}{16}\right)$$ Then the inequality holds for $1< x < x_1$ and for $ x> x2$.
Since the argument is quite small $(-\frac{3 \log (2)}{16}\approx -0.13)$, you can use the series expansions given in the Wikipedia page. For the first root $$W(x)=\sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}x^n =x-x^2+\frac{3}{2}x^3-\frac{8}{3}x^4+\cdots$$ and for the second root $$W_{-1}(x)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots $$ where $L_1=\log(-x)$ and $L_2=\log(-L_1)$.
Using the written terms $$W\left(-\frac{3 \log (2)}{16}\right)\approx -0.15091\implies x_1 \approx 1.16116$$
$$W_{-1}\left(-\frac{3 \log (2)}{16}\right)\approx -3.21341\implies x_2 \approx 24.7252$$ which are quite close to the "exact" solutions.
A: Let us turn the inequality into the equation $\frac{3x}{16} \frac{ \ln(2)}{\ln(x)}=1$. This can be rearranged to
\begin{eqnarray*}
\frac{1}{x} \ln( \frac{1}{x} ) = \frac{-3 \ln 2}{16} 
\end{eqnarray*}
Now let $w=e^{\frac{1}{x}}$ so
\begin{eqnarray*}
we^{w} = \frac{-3 \ln 2}{16} 
\end{eqnarray*}
This can be solved using the Lambert function ... https://www.wolframalpha.com/input/?i=lambert+w(-(3+ln2)%2F16)
\begin{eqnarray*}
w = -0.1511755 \cdots 
\end{eqnarray*}
and $x=1.163201 \cdots$ so $\color{red}{1<x<1.163201 \cdots}$.
EDIT:

https://www.desmos.com/calculator/b7swx2g2xc
This can be solved using the Lambert function ... https://www.wolframalpha.com/input/?i=ProductLog%5B-1,(-(3+ln2)%2F16)%5D
\begin{eqnarray*}
w =W_{-1}( \frac{-3 \ln 2}{16})= -3.20529 \cdots 
\end{eqnarray*}
and $x=24.66272 \cdots$ so $\color{red}{x>24.66272 \cdots}$.
