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Solve for $x$ if $4^x + x^x = 20$ where $x \in \mathbb{Z}$

My Attempt:

By inspection,
$x=2$

Now, $4^2 + 2^2 = 20$

$16+4=20$

$20=20$

Which is True.

But, I Could not do it by any process. Can anyone help me?

Thanks in Advance.

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    Is $x$ integer? – 2017-01-06
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    @barak manos, yes it is. – 2017-01-06
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    Given that the function on the LHS is strictly increasing for $x \geq 1$ then you can try an $x$ and see if it is too large or too small and then try again accordingly. Since $x$ is integer this process will eventually give you an answer. But other than that I would be surprised if a closed-form solution exist. – 2017-01-06
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    If this is actually for a precalculus course, the method they want you to use is 100% inspection and not numerical methods. – 2017-01-06

2 Answers 2

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If the solution is an integer,

$$4^1+1^1<20,\\4^0+0^0<20^*,\\\frac1{4^n}+\frac{(-1)^n}{n^n}\le1+1<20$$ and for all $n>2$, $$4^n+n^n\ge16+4=20.$$

$^*$Whatever the meaning of $0^0$.

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    But what if it was to solve for $4^x + x^x = 16000$? We should check for each integer?(more like trial and error) – 2017-01-06
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    @Ramanujan: this is another question. (Note anyway that $6^6>16000$.) – 2017-01-06
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    @KanwaljitSingh: had the question been asked for arbitrary RHS, I would have answered differently. But it has not. – 2017-01-06
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    @YvesDaoust RHS was arbitrary choose by me " – 2017-01-06
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    @Ramanujan: post a new question if you want to. – 2017-01-06
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    @KanwaljitSingh *All questions of that type* - Which type? What is the definition for it? – 2017-01-06
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    @KanwaljitSingh: please stop the nonsense on this post. – 2017-01-06
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    We are here for teaching and learning also. So its our responsibility if we have some doubt we can ask questions. So that the user who is asking question gets help and better understanding. – 2017-01-06
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    @KanwaljitSingh: it is your responsibility to write comments with a meaning. Also, the link you inserted refers to an equation of a different type and is irrelevant. – 2017-01-06
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    @KanwaljitSingh: it is not a matter of opinion. You shoud realize that your sentences can't be understood. – 2017-01-06
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    @ Yves Daoust, I could not understand from third step. Please elaborate... – 2017-02-07
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    @ Yves Daoust, Could you please explain this solution to.me? – 2017-02-08
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    @PierredeFermat: try with values $\cdots-3,-2,-1$ and $2,3,4\cdots$. – 2017-02-08
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    @ Yves Daoust, please elaborate the second step? (*) – 2017-02-08
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    @ Yves Daoust, I mean why do you write $20^{*}$? – 2017-02-08
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    @PierredeFermat: hem, footnote. – 2017-02-08
2

There does not exist an explicit solution for $x$ which can be expressed in a finite number of elementary functions.

However, you can solve this using an iterative method. In order to apply it, we will generalise our possible solutions to $x \in \mathbb{R}$.

We will use the Newton-Raphson method. The methodology is as follows:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$

We can set a suitable 'guess' in order to obtain a solution to $x$.

For your function, $f(x)=4^x+x^x-20$, we evaluate the derivative to be: $f'(x)=x^x(\ln{x}+1)+\ln{(4)} \cdot 4^x$

We can substitute our functions into equation $(1)$.

$$x_{n+1}=x_n-\frac{4^{x_n}+{x_n}^{x_n}-20}{{x_n}^{x_n}(\ln{x_n}+1)+\ln{(4)} \cdot 4^{x_n}}$$

Let's use an initial value of $x_0=11$. On $x_{29}$, the iteration is accurate to over 10 decimal places (extremely close to $2$). As $n \to \infty$, $x_n \to 2$.

You can apply this iteration into a spreadsheet, or you can use more sophisticated software such as MATLAB.

As others have noted, there exists only one solution to your equation. If there were more solutions, you can set different values for $x_0$ and see if they converge to different solutions.