If (1+x)^2=2^x, what is x?

Question

Show your calculation please.
It strengthens your answer. (Should it be rational)
Thanks!

Answer 1

You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables. The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track. I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).

In this case, I'll use MathCAD. The solutions are:
x={-1.57862, 0, 5.31972} (two of these are approximate, by the way)

Answer 2

There are three real solutions, of which x = 0 is the only rational solution. If we let F[x] = (1+x)^2, G[x] = 2^x, then F[0] = G[0], F[5] > G[5], and F[6] < G[6], so there is a solution in the interval (5,6). Similarly, F[-2] > G[-2] but F[-1] < G[-1], so there is another solution in (-2,-1). Since F and G have derivatives of all orders, and F, G are both convex for all x, it follows that there is are at most (and hence exactly) three solutions. The nontrivial values are found by Newton's method, since the relation cannot be analytically solved in closed form.

Answer 3

As has been said there is no straight forward method. Some
answers here come from deranged people.
First I took the square root of both sides resulting in
x+1=2^(x/2), then
1=2^(x/2)-x, then I set x to 0,1,2,3,4,5,6 in the right side
and got 1,.414, 0,-1,0,.6,2 correspondingly.
So immediately there's a solution at x=0. Checking:
(1+0)^2=1^2=1 and 2^0=1. But since x=5 and x=6 gave
solutions of .6 and 2 respectively which brackets 1, I looked
for a solution between 5 and 6. By tediously bracketing 1
closer and closer I satisfied myself with 5.32 which results
with 1.0003. That is , 2^(5.32/2)-5.32=1.0003+.

Answer 4

0 works but there may be more than one correct answer. Some problems like this require iteration or graphs for a solution. Just graph the right side and the left side vs. x and look for the places where the curves intersect. (Use the BASIC programming language in DOS.) To do it using algebraic calculations would normally mean taking the logarithm of both sides and then expanding it into its "standard series", i.e. an infinite series of terms that gradually diminish, and then simplifying. These are the ways many equations in high order exponents are solved.

Answer 5

Yeah, 0:
(1+0)^2=1
2^0=1
1=1

x=0
:)

Answer 6

x=1

2^2=2^2

Answer 7

take the square root of each side:

(1+x) = 2^(x/2) or -2^(x/2)
so x = 2^(x/2)-1 or -2^(x/2)+1
both roots are 0 so x = 0

check: (1+0)^2 = 1 and 2^0 = 1 so check works

Answer 8

It can be 0
bcoz
substitute 0 in eqn (1+x)^2 = 2^x
then Lhs= (1+0)^2 ...Lhs = left hand side
=1^2
=1
Rhs=2^0 ...Rhs = right hand side
= 1 ... laws of indices(a^0=1)

Answer 9

(1+x)^2=2^x
since they are equal and one side is a square number, the other side must be a square number. therefore x must be even, so that 2^x is a square number.
if x is even, though, 1+x is odd, so (1+x)^2 is also odd. but any power of 2 is going to be even, so the two sides cannot possibly be equal. what is going on?? where is my mistake??

Answer 10

The square root of (1+x)^2 = The square root of 2^x

(1+x) -1 = (The square root of 2^x) -1

x=(The square root of 2^x) -1