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might be trivial for some but I have problems with this equation. I know the answer for $x + 3^x = 4$ is $x = 1$, however I only see that through visual analysis. I searched for an answer on this specific problem online (since it's in the Michael Spivak Calculus book Chapter 1 Problem 4(xii)) and all I was able to find is "by visual analysis... you can see..." which doesn't answer my question

How can you solve this analytically? Thanks in advance

JDF
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Joao Silva
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    I guess you want to solve the equation $x+3^x=4.;$ For this see https://en.wikipedia.org/wiki/Lambert_W_function#Example_1 which handles the case $p^x=ax+b$. – gammatester Sep 13 '16 at 07:06
  • I don't think you can solve it algebraically. But let $f(x) = x+3^x$. $f'(x) = 1 + \ln3*3^x > 0$. So the function is strictly and monotonically increasing. And it's continuous. So every $f(x) =c$ will have exactly one solution for each c in the range of $f$. It's pretty clear all reals are in range so all reals have soultion. $f(0) = 1$ so $f(x) = 4$ has solution at some $x > 0$. $f(2) = 11$ so the solution is in $0 < x < 2$. It doesn't take too much effort to eventually try $f(1) = 4$. – fleablood Sep 13 '16 at 07:10
  • I suppose you meant to use 'equals to' instead of 'less than' in the title... – CiaPan Mar 20 '25 at 13:24

5 Answers5

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From $f'(x)=1+3^x\ln 3>0$, you see that the function is monotonic increasing and cannot have more than one root, which you know to be $x=1$.

Then

$$x<1.$$

2

In general: $$x+a^x=b$$ $$-a^x=x-b$$ $$-e^{x\ln a}=x-b$$ Write $b-x=z$: $$-e^{(b-z)\ln a}=-z$$ $$e^{b\ln a-z\ln a}=z$$ $$e^{b\ln a}=ze^{z\ln a}$$ $$\ln a\cdot e^{b\ln a}=z\ln a\cdot e^{z\ln a}$$ Apply the Lambert W function's defining relation $x=W(x)e^{W(x)}$: $$z\ln a=W(\ln a\cdot e^{b\ln a})$$ $$z=\frac{W(a^b\ln a)}{\ln a}$$ $$x=b-\frac{W(a^b\ln a)}{\ln a}$$

The above derivation is for $x+a^x=b$. To solve $x+a^x<b$, consider the argument to Lambert's W, $w=a^b\ln a$:

  • If $w=-\frac1e$ or $w\ge0$ then $W(w)$ can take on only one value. The corresponding solution to $x+a^x=b$ – call it $x_0$ – is such that the inequality holds for $x<x_0$.
  • If $-\frac1e<w<0$ then $W(w)$ can take on two values, leading to two solutions for $x+a^x=b$: $x_0$ and $x_1$. The inequality holds for $x_0<x<x_1$.
  • Else, $W(w)$ is not defined and the inequality does not hold anywhere.
Parcly Taxel
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Assume $x > 1$ is true; i.e. $1 < x$ and $x - 1 > 0$;

from chapter 17 theorem 4 part (2) (Spivak calculus first edition, 1967); If $x < y \text{ and } a > 1$ then $a^x<a^y$

we have $1 < x$, $\quad$ hence $3 < 3^x$; $\quad$ hence $3^x - 3 > 0$;

hence $(x - 1) +(3^x - 3) > 0; \qquad (x+3^x) -4 >0$;

Hence $x+3^x>4$, which is not the given inequality. Hence assumption $x > 1$ is false.

Take $x = 1$; hence given inequality is $1 + 3^1 < 4;\: 4 < 4$ is false. Hence assumption $x = 1$ is false.

Take $x < 1$; hence $1 - x > 0$ and $3^x<3; \;$ hence $3-3^x>0$;

Hence $(1 - x) + (3-3^x) > 0; \; 4-(x+3^x) > 0; \;$

Hence $x+3^x<4$, which is the given inequality.

Hence $x < 1$ is true and is the solution of given inequality.

Random Commentor
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The thing is, you don't need to solve this analytically; it's easier to state that since the mapping from $x$ to $x+3^x$ is injective, it is also true that for any $a \neq 1$, we have $1+3^1\neq a+3^a$. Therefore the solution you find via visual inspection is the only solution.

H. sapiens rex
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Consider $f(x) = x+3^x$

$f'(x) = 1 + \ln(3)3^x$ on $x \ge 1$

$f'(x) > 0$, so function is growing up. So minimum in $x=1$ (for this set of points). So for other points function will be less than $4$.

$x<1$

Bob Dobbs
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openspace
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  • Also you could use: sum of continuous functions is continuous function – openspace Sep 13 '16 at 07:10
  • Your derivative is wrong. –  Sep 13 '16 at 07:31
  • Yes, the derivative is wrong (most probably a typo), as it should be $;\log 3;$ instead of $;\log x;$ there, but the whole argument is correct. +1 – DonAntonio Sep 13 '16 at 07:53
  • @DonAntonio my bad – openspace Sep 13 '16 at 07:54
  • @DonAntonio: the discussion states $x\ge1$ (so what for $x<1$ ?), then claims a minimum at $x=1$ (?) and concludes that other values won't exceed $4$. I wouldn't call that a typo but a completely misleading answer. –  Sep 13 '16 at 12:52
  • @YvesDaoust That is not my opinion. Certainly some things could be sharpened and written in a smoother way, yet the answer is basically correct imo. I understood from what's written in the answer that :$;f'(x)>0;$ and thus the function grows always (and thus any value the function obtains it is obtained at one unique point). The fact that $;x=1;$ is a minimum is something, I guess, that can be left to the OP. I can't see how this, with all its tiny mistakes, is "completely misleading" ... – DonAntonio Sep 13 '16 at 13:15
  • @DonAntonio: restricting the domain to $x\ge1$ is. You can't justify an answer which is correct by chance. –  Sep 13 '16 at 13:21