# How would you solve this problem using log?

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

$$10n^2 = 2^n$$

• In order to get a good help, it is important to provide your own thoughts for the question. – Arnaldo Mar 22 '17 at 17:15
• 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. – projectilemotion Mar 22 '17 at 17:16
• there are three Solutions expressed by the Lambert-W function – Dr. Sonnhard Graubner Mar 22 '17 at 17:16
• You say you have two equations. I see only one. – Paul Sundheim Mar 22 '17 at 17:18
• One solution is smaller then $10$ since $10\cdot 10^2=1000$ and $2^n=1024$ – kingW3 Mar 22 '17 at 17:23

$y =10n^2$ 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 $9<x_3<10$).