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?

  • 1
    $\begingroup$ 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 $x_{n+1}=x_n-\frac{3x_n-16\left(\frac{\ln{x_n}}{\ln{2}}\right)}{3-\frac{16}{x_n\ln{2}}}$ $\endgroup$ – msgcas May 21 '17 at 22:43
  • $\begingroup$ Thanks. I'll take a look at Newton's method. Approx is fine. $\endgroup$ – 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: enter image description here


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

  • $\begingroup$ Should be $1 < x \color{red}{<} e^{-W\left(-\frac{3}{16}\log 2\right)} \approx 1.163200784... $ and $x > e^{-W_{-1}\left(-\frac{3}{16}\log 2\right)} \approx 24.662720118...$ $\endgroup$ – achille hui May 21 '17 at 22:50
  • $\begingroup$ @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 $\endgroup$ – Donald Splutterwit May 21 '17 at 22:55
  • 1
    $\begingroup$ @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 $1$. $\endgroup$ – achille hui May 21 '17 at 22:58
  • $\begingroup$ @achillehui You are right. I will edit my answer. Thank you for correcting my solution. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.