Solve for the value of k for $(1+\frac{e^k}{e^k+1})^n$

Question

Can anyone find the value of $k$ of the following function: $$(1+\frac{e^k}{e^k+1})^{25} = \frac{3000(\frac{e^k}{e^k+1})-300}{2500(\frac{e^k}{e^k+1})-300}$$

The same equation can also be rewritten as: $$\frac{2500}{3000}=\frac{\frac{\frac{e^k}{e^k+1}-0.1}{(1+\frac{e^k}{e^k+1})^{25}}+0.1}{\frac{e^k}{e^k+1}}$$

Answer 1

For the fun of it.

As Jan Eerland wrote, you look for the zero of function $$f(x)=\left(1+x\right)^{25}-\frac{3000 x-300}{2500 x-300}$$ which has a discontinuity at $x=\frac{3}{25}$.

So, consider instead $$g(x)=({2500 x-300})\left(1+x\right)^{25}-({3000 x-300})$$ and let us make a series expansion around $x=\frac{3}{25}$. This will give $$g(x)=-60+\frac{5613310494340837640403842978331515272 }{142108547152020037174224853515625} \left(x-\frac{3}{25}\right)+O\left(\left(x-\frac{3}{25}\right)^2\right)$$ Ignoring the higher order terms, we then get $$x=\frac{3}{25}+\frac{2131628207280300557613372802734375}{1403327623585209410100960744582878818}\approx 0.121519$$ leading to $$k=-\log \left(\frac{30819917013692599508280802060754974621}{4263273575937635744243216553 816995829}\right)\approx -1.97812$$

Answer 2

Well, we can set:

$$x:=\frac{e^\text{k}}{1+e^\text{k}}\tag1$$

So, we get (for the real solutions):

$$\left(1+x\right)^{25}=\frac{3000\cdot x-300}{2500\cdot x-300}\space\Longleftrightarrow\space x=0\space\vee\space x\approx0.12146650978133148228\tag2$$

I used Mathematica 10.0 to check $\left(2\right)$.

Now, $0$ is not a solution. So now we need to solve:

$$\frac{e^\text{k}}{1+e^\text{k}}\approx0.12146650978133148228\space\Longleftrightarrow\text{k}\approx-1.978615443822769835\tag3$$