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

$10n^2 = 2^n$

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  • $\begingroup$ 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
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    $\begingroup$ there are three Solutions expressed by the Lambert-W function $\endgroup$ – Dr. Sonnhard Graubner Mar 22 '17 at 17:16
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    $\begingroup$ 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
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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 $9<x_3<10$).

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