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?
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 $s=3$.