2
$\begingroup$

I recently stumped across a problem, which I need to solve. Of course, I used an calculator and I got $s=3$, but I want to know how to do it step by step.

The problem is kind of complex: $$\frac{2^{3s+11}-s^5}{(\frac{10s}{3})^5}-\frac{999}{100000s}=10.48$$ Here's what I tried doing:

$$\frac{243(2^{3s+11})-243s^5}{100000s^5}=10.48+\frac{999}{100000s}$$

$$243(2^{3s+11})-243s^5=1048000s^5+999s^4$$

$$2^{3s+11}-s^5=\frac{1048000s^5+999s^4}{243}$$

$$2^{3s+11}=\frac{1048243s^5+999s^4}{243}$$

$$2^{3s}=\frac{1048243s^5+999s^4}{497664}$$

$$2^s=\sqrt[3]{\frac{1048243s^5+999s^4}{497664}}$$

At this point, I tried simplifying the problem in two ways:

  • First by taking logarithms. When I did this way, I ended up in an endless loop where I was breaking down logarithm into parts by logarithm properties, but I had to join them together again and I got back to where I started.

  • Another way I tried, is I tried turning what I got into some kind of quadratic equation somehow, but I didn't really succeed. This is what I did:

$$2^s=\frac{\sqrt[3]{1048243s^5+999s^4}}{\sqrt[3]{13824\,\,\cdot\,\,36}}$$

$$2^s=\frac{s\sqrt[3]{1048243s^2+999s}}{24\sqrt[3]{36}}$$

$$24\sqrt[3]{36}\,\,\cdot\,\,2^s=s\sqrt[3]{1048243s^2+999s}$$

$$\frac{24\sqrt[3]{36}\,\,\cdot\,\,2^s}{s}=\sqrt[3]{1048243s^2+999s}$$

$$\frac{(24^3)(\sqrt[3]{36}^3)(2^s)^3}{s^3}=1048243s^2+999s$$

$$\frac{497664(2^{3s})}{s^3}=1048243s^2+999s$$

When I do it this way, I also get into an endless loop. I also don't know if I can bring everything on one side and set it to zero, and then solve it using quadratic formula. Probably not because one of the terms is $\frac{c}{s^3}$.

How can I then solve for $s$? Am I doing it right, or did I miss something? Remember the original equation is: $$\frac{2^{3s+11}-s^5}{(\frac{10s}{3})^5}-\frac{999}{100000s}=10.48$$

Thank you for help, and please don't vote down for no reason.

EDIT: @callculus commented that this equation can't be solved algebraically, but numerically. How would you solve it numerically?

$\endgroup$
  • $\begingroup$ In general this equation can not be solved algebraically. You have to solve it numerically, for instance by using the newton-raphson method. $\endgroup$ – callculus Mar 31 '16 at 16:13
1
$\begingroup$

First of all, we can observe $3s$ must be an integer, in other cases the first fraction would be irrational, so suppose $s=\dfrac{a}{3}\quad ,a\in\mathbb Z^+$

$$\frac{2^{3s+11}-s^5}{(\frac{10s}{3})^5}-\frac{999}{10^5s}= \frac{9^5\cdot2^{a+11}-3^5\cdot a^5}{10^5\cdot a^5}-\frac{3\times999}{10^5a}=10.48$$

$\Rightarrow 9^5\cdot2^{a+11}-3^5\cdot a^5-3\times999a^4=1048000\times a^5$

Now we can see $a\mid9^5\cdot 2^{a+11}$, so $a=3^\alpha2^\beta$ $$9^5\cdot 2^{3^\alpha2^\beta+11}-3^5\cdot 3^{5\alpha}2^{5\beta}-3\times999(3^{4\alpha}2^{4\beta})=1048000\times 3^{5\alpha}2^{5\beta}$$ And if we divide both sides by $2^{4\beta}$ we will see $\beta=0$, so a=$3^\alpha$

Now observe that $a$ is not a big integer, since $2^{a+11}$ could be so huge: $$1048000\times a^5+3^5\cdot a^5+3\times999 a^4\lt2\times10^6a^5$$ For instance, for $a=20$ $$2\times10^6a^5=400\times10^{9}\lt400\times1024^3\lt9^5\cdot 2^{31}$$ So $a$ is a power of $3$ which must be under $20$, therefore the set of answers for $s=\dfrac{a}{3}$ is $\{1,3\}$. And it only holds for $s=3$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't think your first sentence is correct if $s$ isn't assumed to be rational. The overall expression could be rational even if each of the pieces is irrational. In fact, the intermediate value theorem requires the overall expression to be rational on a dense set of inputs. $\endgroup$ – Ravi Fernando Mar 31 '16 at 18:52
  • $\begingroup$ You are right, let me see if I can correct this solution or not. $\endgroup$ – Amir Naseri Mar 31 '16 at 19:05

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.