# Solve the equation with respect to $t$

How can I solve this equation with respect to $t$:

$$b(2^{\ln (t)/\ln(2)+2}-1)a^{t}-a^{2t}-1=0$$ where $a$ and $b$ are real numbers.

• Does $2^{((ln(t))/(ln(2)))+2}$ read $2^{\frac{\log(t)}{\log(2)}+2}$ ? – Claude Leibovici Oct 20 '14 at 9:16
• @ClaudeLeibovici: Yes, it is. – DER Oct 20 '14 at 9:19
• Have you typed the equation in correctly? As it is now, it simplifies down to $b(4t-1)a^t-a^{2t}-1=0$, which I can't find a solution for. – mardat Oct 20 '14 at 9:23
• @mardat: How do you get this last form! – DER Oct 20 '14 at 9:26
• Note that $ln(t)/ln(2) = log_2{t}$, which means that $2^{ln(t)/ln(2)+2} = 2^2*2^{log_2{t}}$, and since $a^{log_a{t}} = t$, it simplifies further to $2^2*t = 4t$. Both of those are properties of log. – mardat Oct 20 '14 at 9:28

After your clarification to my comment, you could notice (using logarithms) that $$2^{\frac{\log(t)}{\log(2)}+2}=4t$$ So, your equation becomes $$b(4t-1)a^t-a^{2t}-1=0$$ that is to say $$b= \frac{a^t+a^{-t}}{4t-1}$$ provided $t \neq\frac{1}{4}$. Rewriting $a^t=e^{t\log(a)}$, then the equation is $$b=\frac{2\cosh(t\log(a))}{4t-1}$$ and I do not think you could go much further analytically. At this point, root-finders (such as Newton method) would solve the problem.
For given $a$ and $b$, you could notice that you look for the intersection of the straight line $$f(t)=b(4t-1)$$ and the curve $$g(t)=2\cosh(t\log(a))$$
Since $a=(√3+2)^{2^{p-2}}$, $p$ being a prime, let us choose $p=5$ and $b=123456789$. A plot of the function $$F(x)=b(4t-1)-2\cosh(t\log(a))$$reveals that there is one root close to $t=2$. So, let us use Newton method which, starting with a guess $x_0=2$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Applied to the problem, the successive iterates will ten be $1.96173$, $1.95107$, $1.95038$ which is the solution for six significant figures.