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I have the following exponential equation: $2^x = 5 + x$ Can anyone help to solve it?

With Thanks

  • 0
    a plot of the function $$f(x)=2^x-5-x$$ Shows us that we get two real solutions – 2017-02-08
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    By inspection, $x=3$. You need to check if this is the only solution. – 2017-02-08
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    It has exactly two solutions. Luckily the positive one is $3$. The negative one is a rather anonymous real number, especially if you do not use the Lambert-W function. – 2017-02-08
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    There does not exist a closed form solution for $x$ in terms of elementary functions. Try using the [Newton-Raphson method](https://en.m.wikipedia.org/wiki/Newton's_method). – 2017-02-08

1 Answers 1

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Taking the derivative of $2^x-x-5$,

$$\ln2\,2^x-1,$$ reveals the existence of a minimum, at $x=-\dfrac{\ln(\ln2)}{\ln2}$. As the ordinate of this minimum is negative, there are two roots.

By inspection, one is $x=3$ and we can infer that there must be one close to $x=-5$, as $2^{-5}$ is small.

After one iteration of Newton's method, we get

$$x\approx-5-\frac1{\ln 2-32}=-4.9680581115653\cdots$$

It is such that $f(x)<10^{-5}$.