<s>[INST] Solving logarithm problem, Found the value This is my first question in this place, I don't know how to solve my problem.
I have this equation, I need to find the central value based on this equation:
$k = 0.2$
$$k = 2^{(\frac{1}{24} + \frac{1}{24\cdot2^x})}-2^{(\frac{1}{24} - \frac{1}{24\cdot2^x})}$$
I need to figure out the x value, the known value is k.
$$x=?$$
if I define y like: $y = \frac{1}{24\cdot2^{x}}= 2^{-x}/24$,  and l like:  $l = k/2^{\frac{1}{24}}$
$$k = 2^{\frac{1}{24}}\cdot2^y-2^{\frac{1}{24}}\cdot2^{-y}$$
$$l = 2^y-2^{-y}$$
$$\log_{2}(l)=\text{ ?}$$
$$y=\text{ ?}$$
Please help me.
 [/INST] First of all, set $a=1/24$ and $t=2^{-x}$, so the equation becomes
$$
k=2^{a+at}-2^{a-at}
$$
We now see a further simplification: set $b=2^a$, so the equation becomes
$$
k=b^{1+t}-b^{1-t}
$$
or else
$$
k=bb^t-\frac{b}{b^t}
$$
that can be rewritten as
$$
bb^{2t}-kb^t-b=0
$$
This is a quadratic in $b^t$ and we can immediately discard the negative root, so
$$
b^t=\frac{k+\sqrt{k^2+4b^2}}{2b}
$$
Therefore
$$
t=\frac{\log(k+\sqrt{k^2+4b^2})-\log b-\log 2}{\log b}
$$
Recalling that $t=2^{-x}$ we get
$$
-x\log 2=\log\frac{\log(k+\sqrt{k^2+4b^2})-\log b-\log 2}{\log b}
$$
The logarithm can be with respect to any basis, the formula is simpler if we use base $2$, so $\log_2b=\log_22^a=a$ and $\log_22=1$, so
$$
-x=\log_2\frac{\log_2(k+\sqrt{k^2+4b^2})-a-1}{a}
$$
Since $b=2^a$ we have $b^2=2^{2a}$ and $4b^2=2^{2+2a}$ and we have
$$
x=\log_2\frac{a}{\log_2(k+\sqrt{k^2+2^{2+2a}})-a-1}
$$
</s>