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.

  • $\begingroup$ Does $2^{((ln(t))/(ln(2)))+2}$ read $2^{\frac{\log(t)}{\log(2)}+2}$ ? $\endgroup$ – Claude Leibovici Oct 20 '14 at 9:16
  • $\begingroup$ @ClaudeLeibovici: Yes, it is. $\endgroup$ – DER Oct 20 '14 at 9:19
  • $\begingroup$ 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. $\endgroup$ – mardat Oct 20 '14 at 9:23
  • $\begingroup$ @mardat: How do you get this last form! $\endgroup$ – DER Oct 20 '14 at 9:26
  • $\begingroup$ 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. $\endgroup$ – 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 $$g(t)=2\cosh(t\log(a))$$

Added later

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.