Do you think value of n related with Euler constant (e) ?

Question

0.5^n = n
n = ?

Answer 1

gg - complex numbers involve i, not e particularly.

I don't see any reason why n would be particularly related to e. If we write n = 2^r we get (2^-n) = n => 2^(-2^r) = 2^r and hence r = - 2^r. Since 2^r is always positive we know r is negative, and since r is negative 2^r is between 0 and 1. So r is between -1 and 0 (and hence n = 2^r is between 0.5 and 1). But there's not going to be any "nice" solution for a problem of this form, all you can do is get one numerically. In this case I get n = 0.641186 to 6 d.p.

Answer 2

Take logarithms of both sides in base 1/2, then you get
n=log_(0.5)n, or if you prefer intgers
n=-log_2 n which you can also write as n+log_2 n=0.

Now consider the function f(x)=x+log_2 x which is a continuous function. Use a calculator if you like, to check that f(1/2)=1/2-1=-1/2 and on the other hand f(1)=1+0=1. Therefore in between 1/2 and 1 there must be a point where f(x)=0 . At that point you have your solution and it is definitely a positive real number. By the way I don't think there is an analytic way to find that solution itself, probably you need to use a numeric method or a calculator to find the exact value.

Answer 3

no . its not a complex no . n is real no