Solving 1= \frac{2^{x+1}}{x}: Is There a Trick?

[cragar]

Homework Statement

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

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?

 

[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.

 

[Ray Vickson]

Homework Statement

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

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?

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

 

[Ocasta]

Homework Statement

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

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.

 

[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))

 

[cragar]

ok thanks for the answers

 

[Mark44]

I think there is no real solution.

Since this question was posted in the Precalc section, I'm going to assume that the domain for consideration is real numbers. In that case, I agree with ritoban5 that there is no real solution, and in that context, that would be the answer.

cragar,
Is there some reason we should consider the complex numbers as possible solutions?