1. ## Simple equation help?

Hi there,

This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge:

2^x = -x

2. ## Re: Simple equation help?

You will need to solve it numerically, guess and check or technology.

Graph $y = 2^x$ & $y= -x$ , to get a rough idea on where they intersect.

3. ## Re: Simple equation help?

Originally Posted by eskimogenius
Hi there,

This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge:

2^x = -x
The equation You have written is trascendental and its solution has to be found numerically. A non well popular but elegant way to solve it is to writre the equation as $f(x)= -2^{x}-x=0$ [th sign '-' has a precise scope...] and then compute for n large enough the n-th term of the sequence defined by the 'recursive relation' ...

$x_{n+1}=-2^{x_{n}}\ ;\ x_{0}=0$ (1)

In few step You arrive at the solution $x \sim -.641186$. The reason for which $x_{n}$ tends to the [real] solution of Your equation will be explained in a tutorial post written in the section 'Discrete mathematics'...

Kind regards

$\chi$ $\sigma$

4. ## Re: Simple equation help?

Originally Posted by chisigma
The equation You have written is trascendental and its solution has to be found numerically. A non well popular but elegant way to solve it is to writre the equation as $f(x)= -2^{x}-x=0$ [th sign '-' has a precise scope...] and then compute for n large enough the n-th term of the sequence defined by the 'recursive relation' ...

$x_{n+1}=-2^{x_{n}}\ ;\ x_{0}=0$ (1)

In few step You arrive at the solution $x \sim -.641186$. The reason for which $x_{n}$ tends to the [real] solution of Your equation will be explained in a tutorial post written in the section 'Discrete mathematics'...

Kind regards

$\chi$ $\sigma$
Sorry, I have no idea what this means, or how to do this.

I have graphed the lines, and realised that they intersect between 0 and -1, however, I don't know how to arrive at the solution.

Can it be done using simple log laws?

5. ## Re: Simple equation help?

the solution is probably transcendental. Do you know newton's method?

6. ## Re: Simple equation help?

Originally Posted by eskimogenius
Sorry, I have no idea what this means, or how to do this.

I have graphed the lines, and realised that they intersect between 0 and -1, however, I don't know how to arrive at the solution.

Can it be done using simple log laws?
The equation...

$f(x)=- 2^{x}-x=0$ (1)

... cannot be solved using simple log laws!... once You have found in graphical way that the solution is between -1 and 0, if You need better accuracy the only choice You have are numerical methods. Among them one of the most simple [and elegant...] even if not widely used is to consider the solution as the limit [if it exists...] of the sequence defined by the recursive relation...

$\Delta_{n}= x_{n+1}-x_{n}= f(x_{n})$ (1)

In Your case the (1) is...

$x_{n+1}= -2^{x_{n}}$ (2)

What You have to do now is to compute iteratively the $x_{n}$ till to an n 'large enough'. Starting with $x_{0}=0$ You easily find...

$x_{1}= -2^{0}= -1$

$x_{2}= -2^{-1}= -.5$

$x_{3}= -2^{-.5}= -.707106...$

$x_{4}= -2^{-.707106...}= - .612547...$

$x_{5}= -2^{-.612547...}= -.65404...$

$x_{6}= -2^{-.65404...}= -.6354978...$

$x_{7}= -2^{-.6354978...}= -.6437186...$

$x_{8}= -2^{-.6437186...}= -64006102...$

... and so one. The convergence to the solution $x \sim -.641186$ is evident...

Kind regards

$\chi$ $\sigma$

7. ## Re: Simple equation help?

Originally Posted by eskimogenius
Hi there,

This popped up in my exam the other day, absolutely no idea how to solve it using secondary school knowledge:

2^x = -x
Clearly you are not expected to solve it by hand, for the simple fact that it cannot be solved exactly by hand or otherwise (unless you use a special function called the Lambert W-function).

Did you have a graphics or CAS calculator in the exam? You are expected to know how to use your calculator to solve it.
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