# 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?

• I don't now how to solve this enequality exactly, but you can try to approximate the solutions of the equality using Newton's method: Choose a start value $x_0$ and iterate $x_{n+1}=x_n-\frac{3x_n-16\left(\frac{\ln{x_n}}{\ln{2}}\right)}{3-\frac{16}{x_n\ln{2}}}$ – msgcas May 21 '17 at 22:43
• Thanks. I'll take a look at Newton's method. Approx is fine. – Larry Freeman May 21 '17 at 22:47

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 $\color{red}{x>24.66272 \cdots}$.

• Should be $1 < x \color{red}{<} e^{-W\left(-\frac{3}{16}\log 2\right)} \approx 1.163200784...$ and $x > e^{-W_{-1}\left(-\frac{3}{16}\log 2\right)} \approx 24.662720118...$ – achille hui May 21 '17 at 22:50
• @achillehui .. this value of $x=1.16 \cdots$ gives $1$ when I plug it in my calculator ... I will investigate further. ... Ref for Larry en.wikipedia.org/wiki/Lambert_W_function – Donald Splutterwit May 21 '17 at 22:55
• @DonaldSplutterwit you get the boundary right but the function $\frac{3x}{16}\frac{\log 2}{\log x}$ is decreasing on $(1,e)$ and increasing on $(e,\infty)$. For $x > 1$, there are two ranges where the function is greater than $1$. – achille hui May 21 '17 at 22:58
• @achillehui You are right. I will edit my answer. Thank you for correcting my solution. – Donald Splutterwit May 21 '17 at 23:03

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.