Does it follow that for $ x\ge 785$,

$$(2x+3)^{\frac{1.25506}{\ln x}} < 4$$

Note: $1.25506$ is taken from this inequality regarding the prime counting function $\pi(x)$:

Here's my thinking:

(1) $(2(785+3)^{\frac{1.25506}{\ln 785}} < 4$

To complete the argument, I need to show that for $x \ge 785$, $(2x+3)^{\frac{1.25506}{\ln x}}$ is decreasing.

(2) $(2x+3)^{\frac{1.25506}{\ln x}}$ is decreasing if and only if $\dfrac{1.25506}{\ln x}\ln (2x+3)$ is decreasing so let:

$$f(x) = \dfrac{1.25506}{\ln x}\ln (2x+3)$$

(3) Using the Quotient Rule with $g(x) = 1.25506\ln (2x+3)$ and $h(x) = \ln x$:

$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$

(4) Using the Product Rule with $s(x) = 1.25506$ and $t(x)=\ln(2x+3)$

$$g'(x) = s'(x)t(x) + s(x)t'(x)$$

(5) Using the Chain Rule for $t(x)$ with $q(x) = 2x+3$ and $p(x) = \ln(q(x))$:

$$t'(x) = p'(q(x))'q(x) = (\ln(2x+3))'(2x+3)' = \frac{2}{2x+3}$$

(6) Applying Step(5):

$$g'(x) = 1.25506\frac{2}{2x+3} = \frac{2.51012}{2x+3}$$

(7) Applying Step(4):

$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} = \frac{\frac{2.51012\ln x}{2x+3} - \frac{1.25506\ln (2x+3)}{x}}{(\ln x)^2}$$

(8) The derivative is negative when:

$$\frac{1.25506\ln (2x+3)}{x} > \frac{2.51012\ln x}{2x+3}$$

Restated as:

$$(2x+3)\ln (2x+3) > 2x\ln x$$

which is true for $x > 0$

Is this argument valid? Did I ma 1.25506\ln (2x+3)ke a mistake or present an incomplete argument? In my reasoning in Step(9) sufficient?

Edit: I have restated the question with help from the commenters

Even though I have accepted an answer, please post comments if you see any mistakes or have suggestions for improvements.

  • $\begingroup$ If you want to show that $F(x)=(2x+3)^{\frac{1.25506}{\ln x}}$ is decreasing, why are you considering the derivative of $\frac x{F(x)}$? $\endgroup$
    – saulspatz
    May 18, 2020 at 22:44
  • $\begingroup$ It certainly looks wrong to me. $\endgroup$
    – saulspatz
    May 18, 2020 at 22:48
  • $\begingroup$ I believe that I messed. I was trying to show that $x$ gets larger as a ratio. $\endgroup$ May 18, 2020 at 22:48
  • $\begingroup$ Thanks. I will fix it. I think that I got a head of myself. :-( $\endgroup$ May 18, 2020 at 22:49
  • $\begingroup$ It would be easier to show that $\ln (f(x)) = c\frac{\ln (2x+3)}{\ln x}$ is decreasing using the quotient rule to show that the derivative is negative. $\endgroup$
    – Anand
    May 18, 2020 at 23:11

3 Answers 3


Solution without derivatives.

We need to prove that $$(2x+3)^{1.25506}<4^{\ln{x}}$$ or $$e^{\ln4\ln{x}}-(2x+3)^{1.25506}>0$$ or $$x^{\ln4}-(2x+3)^{1.25506}>0$$ or $$x^{\ln4-1.25506}-\left(2+\frac{3}{x}\right)^{1.25506}>0,$$ which is true because $$x^{\ln4-1.25506}-\left(2+\frac{3}{x}\right)^{1.25506}\geq785^{\ln4-1.25506}-\left(2+\frac{3}{785}\right)^{1.25506}\approx0.0058>0.$$ Now we see that our inequality is true even for any $x\geq771.$


Making the problem a bit more general, I should consider that we need to find the zero of function $$f(x)=\frac{a }{\log (x)}\log (2 x+3)-\log (b)$$ Expanded as a Taylor series, we have $$f(x)=\left(\frac{a \log (2)}{\log (x)}+a-\log (b)\right)+\frac{3 a}{2 x \log (x)}+O\left(\frac{1}{x^2}\right)$$ the solution of which being given in terms of Lambert function $$x=-\frac{3 a}{2 (a-\log (b))} \frac 1{ W\left(-\frac{a}{a-\log (b)} k^{-\frac{a}{a-\log (b)}}\right)}\qquad \text{where} \qquad k=\frac 13 \left(\frac{3}{2}\right)^{\frac{\log (b)}{a}}$$

Using your constant this gives as an estimate $x_0=770.873$. Using Newton method, the iterates for $f(x)=0$ would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 770.8726304 \\ 1 & 770.8589458 \\ 2 & 770.8589459 \end{array} \right)$$


If we use only the first term of the expansion $$\frac{a \log (2)}{\log (x)}+a-\log (b)=0 \implies x=2^{-\frac{a}{a-\log (b)}}$$ which, using your numbers, gives as a simple estimate $x=756.6600$ which is not bad.

It is also interesting to notice the very strong dependency of $x$ to $b$. Using the original equation, be have that $$\frac{db}{dx}=\frac{a \left(\frac{2}{2 x+3}-\frac{\log (2 x+3)}{x \log (x)}\right) (2x+3)^{\frac{a}{\log (x)}}}{\log (x)}$$ which around $x=771$ gives $$\frac{db}{dx}\approx -1.04\times 10^{-4}$$

This probably explains why we need quite accurate calculations.

To illustrate the need of rather accurate calculations $$\left( \begin{array}{cc} x & b \\ 760 & 4.00114 \\ 765 & 4.00061 \\ 770 & 4.00009 \\ 775 & 3.99957 \\ 780 & 3.99905 \\ 785 & 3.99854 \\ 790 & 3.99804 \\ 795 & 3.99753 \\ 800 & 3.99703 \end{array} \right)$$


Just penciling out the ideas in the comments. Let $f(x)$ be your function: $$ f'(x) = \frac{(2 x+3)^{1.25506/\log(x)}}{\log(x)^2}\left(\frac{2.51012\log(x)}{2x+3} - \frac{1.25506 \log(2x+3)}{x}\right) $$Tedious arithmetic shows the second term in parentheses has no real solution and $f'(1)\approx -2.02$. Then $f'$ is always negative, and another calculation shows $f(785)\approx 3.99854$. So yes, eventually $f(x)<4$.


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