Solving logarithm problem, Found the value

Question

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.

Answer 1

Multiplying both side of $l=2^{y}-2^{-y}$ by $2^{y}$ you will get $$ A^{2}-Al-1=0, $$ where $A=2^{y}$. Solving this equation with respect to $A$ gives the desired values for $A$.

Answer 2

$$k=2^{\frac{1}{24}+\frac{1}{24\cdot2^x}}-2^{\frac{1}{24}-\frac{1}{24\cdot2^x}}\Longleftrightarrow$$ $$k=2^{\frac{2^{-x}+1}{24}}-2^{\frac{1-2^{-x}}{24}}\Longleftrightarrow$$ $$k=2\cdot2^{\frac{1}{24}}\sinh\left(\frac{2^{-3-x}\ln(2)}{3}\right)\Longleftrightarrow$$ $$\frac{k}{2\cdot2^{\frac{1}{24}}}=\sinh\left(\frac{2^{-3-x}\ln(2)}{3}\right)\Longleftrightarrow$$ $$\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)=\frac{2^{-3-x}\ln(2)}{3}\Longleftrightarrow$$ $$3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)=2^{-3-x}\ln(2)\Longleftrightarrow$$ $$\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}=2^{-3-x}\Longleftrightarrow$$ $$\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}=-3-x\Longleftrightarrow$$ $$\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}+3=-x\Longleftrightarrow$$ $$x=-\left[\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\cdot2^{\frac{1}{24}}}\right)}{\ln(2)}\right)}{\ln\left(2\right)}+3\right]\Longleftrightarrow$$ $$x=-\frac{\ln\left(\frac{3\text{arcsinh}\left(\frac{k}{2\sqrt[24]{2}}\right)}{\ln(2)}\right)}{\ln(2)}-3\Longleftrightarrow$$ $$x=-\frac{\ln\left(\frac{24\text{arcsinh}\left(\frac{k}{2\sqrt[24]{2}}\right)}{\ln(2)}\right)}{\ln(2)}$$

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EDIT:

In general:

$$2^{\frac{a+1}{x}}-2^{\frac{1-a}{x}}=2^{\frac{1-a}{x}}\left(4^{\frac{a}{x}}-1\right)$$

Answer 3

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 $$ x=\log_2\frac{a}{\log_2(k+\sqrt{k^2+2^{2+2a}})-a-1} $$