Solving 3^x+x^3=17, a Non-Standard Equation

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[angelmendez-rivera]

Let f, g, h : R —> R, g(x) = 3^x everywhere, h(x) = x^3 everywhere, f = g + h. g is isotonic and continuous, and h is isotonic and continuous. Therefore, f is isotonic and continuous. Therefore, f is injective. f is also surjective, since it has no global minimum or maximum. Thus, f is bijective, and so, invertible. As such, f^(–1) exists, and so, provided the equation f(x) = 17 for some real x, x = f^(–1)(17) necessarily. Now, f(x) = g(x) + h(x) everywhere, so g(f^(–1)(17)) + h(f^(–1)(17)) = 17. However, g(x) >= 0 everywhere, so h(f^(–1)(17)) =< 17. Since h(x) = x^3 everywhere, we have f^(–1)(17)^3 =< 17, which means that it is worthwhile to investigate the perfect cubes greater than 0 that are less than or equal to 17. These are 1 and 8. If neither works, then that means there is no closed form for f^(–1)(17), and so numerical approximations must be used. If h(f^(–1)(17)) = 1, then g(f^(–1)(17)) = 3^f^(–1)(17) = 16. However, this would mean that ln(16)/ln(3) = 1, which is false. If h(f^(–1)(17)) = 8, then g(f^(–1)(17)) = 9, thus f^(–1)(17)^3 = 8 and 3^f^(–1)(17) = 9. It turns out, both of these result in f^(–1)(17) = 2. Therefore, x = 2.

EDIT: Actually, we can do better than this. I said it is worthwhile to investigate the positive perfect cubes less than 17, but I never actually proved that this is worthwhile. The function f is transcendental, and its inverse is also transcendental, and furthermore, non-elementary, which can be proven using Galois theory, (but which is obviously outside the scope of my comment). Not even the Lambert W function is enough to make f^(–1) expressible in terms of elementary functions. This means that, unless x = f^(–1)(17) is expressible in terms of radicals, there is no closed form for x, and so there is nothing else to be done with this problem, except to pull out your computers and start approximating. As such, all we must do is actually confirm whether x does have a closed form or not, and this is easy to test. How? Because as noted above, if x does have a closed form, then it must be algebraic. However, this means it must be rational: otherwise, 3^x is transcendental, and x^3 is algebraic, so their sum is transcendental, and by consequence, not equal to 17. Since x is rational, x^3 is rational, so 3^x must be rational if their sum is to equal 17. But 3 is a prime number, so 3^x is rational only if x is an integer. Since x must be an integer, x^3 must be an integer, so this must mean 3^x is an integer, which requires x >= 0. This means x^3 >= 0. So, x^3 must be an integer, and we know 0 =< x^3 < 17. Therefore, x^3 = 0, x^3 = 1, x^3 = 8. If neither of these is true, then x has no closed form. Q. E. D.

[thomy]

Solution by observation: x=2 and the function is increasing so its only solution

[gdtargetvn]

You can also "avoid" using derivatives by proving there's no other solutions other than x = 2.
For x > 2, LHS > RHS.
For x < 2, LHS < RHS.
which means x = 2. (also include a verification step after that)

[gdtargetvn]

y = f(x) = 3^x + x^3 is strictly increasing because f'(x) = 3^x log(3) + 3x^2 > 0, so each x value only gives one y value, which means we can guess.
f(2) = 17 so x = 2 is a unique solution.

[Packerfan]

Define f:R -> R by (x) = 3^x + x^3 - 17.

Note that f is differentiable and continuous on (-inf, inf) and f(2) = 0

f '(x) = 3^x ln 3 + 3x^2 > 0 on (-inf, inf)

Thus, f is increasing and has at most one solution. Because if there is a in R such that f(a) = 0 and a != 2, then by the Mean Value Theorem, there exists c in (a, 2) if a < 2 or c in (2, a) if a > 2 such that f '(c) = 0 since f(a) = f(2) = 0. But this contradicts f '(x) > 0 on (-inf, inf). Thus, f has exactly one root at x = 2.

[ACertainMan]

Solution by physics
Try 1: 4
Try 2: 17

[irinamladova]

My solution:
f(x)=g(x)
f(x)=3^x, g(x)=17-x^3
f increase, g decrease
1 point=>1 answer
See x=2

[stevenlitvintchouk]

Rather than all these nonstandard problems with guess-and-check solutions, perhaps it would be worth one or two videos to look at numerical methods for finding possible roots of nonstandard equations. In this case, even a relatively dumb solver like the secant method converges to the correct answer, x = 2, in just 3 steps. Since it's a numerical method, you still have to verify that 2 is the exact answer by substitution in the original equation. But it's a more systematic approach than just guessing your way along.

[michaelempeigne]

i like to use the largest powers of both expressions :

3^2 = 9 and 3^3 = 27
2^3 = 8 and 3^3 = 27

so 17 < or = 3^x + x^3 < or = 54
since 3^x + x^3 = 17; this requires x = 2 as shown.

[devondevon]

X=2
3^x + x^3 = 3^2 +2^3 since 2^3 +3^2= 17
once you break down 17 into 9 and 8 you notice that 2^3 3^2 exponent becoming base pattern
from inspection x =2 answer

[neilgerace]

I just used guess and check ... Not rigorous but effective this time :)

[lilalila]

You dont need to move x^3 to the RHS. The derivative of the LHS is 3^x.ln3+3x^2, which is always greater than 0, so the LHS is an increasing function. Considering that the RHS is a constant, there is only one solution. Guess and check we have x=2

[reelreflections-us]

Would really like a more general approach h to solving this.

In this case, range limiting and the answer being a small integer worked (common in contest problems), but what if the answer was a fractional number or irrational?

Perhaps a numerical approach using Newton’s method or something like that?

[SuperYoonHo]

Woah! That looked crazy!!! But i did manage to solve it at the very end and surprisingly your solution was same as mine! *Some people think the same...

[j.r.]

I'm pleased that this time I was correct in assuming that there would be no way to calculate the solution algebraically, even though the solution itself was obvious. But your table of values did at least confer some dignity on the process of guessing the answer. :)

[VY_Canis_Majoris]

Well both 3^x and x^3 are always increasing so the only solution is 2.

[ggiieerroo]

That was nice - now let's try another one: x^3 - 3^x = 0 (with two solutions) :)

[christianthomas]

it seems to me that x=2 is the unique solution, obviously.

[Atrulion]

Lmao I just thought that 2 to the power of 3 is 8 and that is already half of 17, so i took 3 to the power of 2 and got 9, meaning they were 17, and that was my solving technique

[vizart]

Isnt these problems a tad too much on the easy side? It boils down to observing a value that works and then prove that its the only one.