# Solve for x

1. Sep 4, 2011

### cragar

1. The problem statement, all variables and given/known data

$1= \frac{2^{x+1}}{x}$
3. The attempt at a solution
I multiplied both sides by x and then tried taking the ln of both sides but it just seems like you go in circles. Is this solvable with some trick?

2. Sep 4, 2011

### LCKurtz

Equations like that can't normally be solved by algebraic methods. You must use numerical methods. However, this particular equation can be shown not to have any real solutions.

3. Sep 4, 2011

### Ray Vickson

It's not straightforward. It is solvable in terms of the so-called Lambert W-function. The solution is complex. Here it is in Maple 14:

eq:=2^(x+1)/x=1;
(x + 1)
2
eq := -------- = 1
x
sol:=solve(eq,x); evalf(sol);
sol := -1/ln(2)*LambertW(-2*ln(2))
.1279196207-2.181686754*I (I = sqrt(-1) in Maple terminology)

Alternatively: you could write x = a + I*b and get a set of two coupled nonlinear equations for a and b, then solve them numerically using a Newton-Raphson method or something similar.

RGV

4. Sep 4, 2011

### Ocasta

You're right about going around in circles.

$(x) 1= \frac{2^{x+1}}{x} (x)$

$x= 2^{x+1}$

$\ln(x)= \ln(2^{x+1})$

$\ln(x)= x \ln(2) + \ln(2)$

$\ln(x) - \ln(2) = x \ln(2)$

$\frac{\ln(x)}{\ln(2)} - 1 = x$

This will give you an x = something, but you still have x in both sides of the equation.

5. Sep 4, 2011

### ritoban5

I think there is no real solution. When I plot x and 2(x+1).....they do not intersect. However if x is imaginary, you get one extra d.o.f and a family of solutions exist.

Suppose x=a+ib. Then your equation becomes two equations (one for Re, and one for Im), these are:

a=2(a+1)(cos(b ln(2))) and b=2(a+1)(sin(b ln(2))). The ratio of these gives a=b/tan(b ln(2)). Thus the family of solutions is x= b/tan(b ln(2))+ib with |b|< $\pi$/(2 ln(2))

6. Sep 4, 2011