I have this equation and I want to find the possible values of $n$. So how would you solve this using logarithms?

$10n^2 = 2^n$

  • $\begingroup$ In order to get a good help, it is important to provide your own thoughts for the question. $\endgroup$ – Arnaldo Mar 22 '17 at 17:15
  • $\begingroup$ You can express the solution in terms of the Lambert W function. Alternatively, if you want to avoid this then you can solve this numerically. For example, you can use the Newton-Raphson method. $\endgroup$ – projectilemotion Mar 22 '17 at 17:16
  • 1
    $\begingroup$ there are three Solutions expressed by the Lambert-W function $\endgroup$ – Dr. Sonnhard Graubner Mar 22 '17 at 17:16
  • 3
    $\begingroup$ You say you have two equations. I see only one. $\endgroup$ – Paul Sundheim Mar 22 '17 at 17:18
  • $\begingroup$ One solution is smaller then $10$ since $10\cdot 10^2=1000$ and $2^n=1024$ $\endgroup$ – kingW3 Mar 22 '17 at 17:23

There is no way to find the solution of this equation using logarithm.

You can find the roots with some numerical method or using the Lambert W function. If you want a simple estimate of the roots tha you can sketch the graphs of the functions

$y =10n^2$ and $y=2^n$ ( it is easy)

and search the common points. You see immediately that there is a negative solution $x_1<0$ and a positive solution $0<x_2<1$ because $10\cdot 1^2 > 2^1$. Since we have $2^{10} < 10\cdot 10^2$ there is also another solution $1<x_3<10$ (and you can easily restrict the interval to $9<x_3<10$).


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.