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I want to mention that it is not my homework, just want to solve for fun. I appreciate any hint how to solve it. The exponential equation is given:

$2^x + 3^x = 10000$

My initial thought was to use such transformation: $2^x + 2^{\log_2{3^x}} = 10000$, but it gives nothing for me

MrYouMath
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Alex Aparin
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  • Did you have a look at this similar question here? – Dietrich Burde Jan 10 '18 at 09:09
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    I think this problem does not have a closed-form solution. – MrYouMath Jan 10 '18 at 09:09
  • @DietrichBurde, it is one of most popular question, but I think it does not have any relation with my equation – Alex Aparin Jan 10 '18 at 09:10
  • @MrYouMath, Why do you think so? Maybe there is any theorems about that? – Alex Aparin Jan 10 '18 at 09:11
  • First of all, note that $2^x+3^x$ is an increasing function and $2^0+3^0=2<10000$, while $\lim_{x \rightarrow \infty} 2^x+3^x=\infty$, hence you have exactly $1$ soultion by using IVT. You can even approximate it to be between $8$ and $9$ but it still looks ugly to me – asdf Jan 10 '18 at 09:11
  • But the methods could be used. – Dietrich Burde Jan 10 '18 at 09:11
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    @LmTinyToon: I do not know any theorem that allows us to directly draw this conclusion. This is more of my intuition. The equation looks simple, but with the sum of two different exponentials, it is most likely not possible. I am sure there are infinite expressions for something like this (e.g. I think you could use fixed point iteration to get such a solution) . – MrYouMath Jan 10 '18 at 09:23
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    @MrYouMath I agree, seems that the best try is to make an approximation... – Macrophage Jan 10 '18 at 09:35
  • Wolfram alpha does not find a closed-form-solution. It seems that even the Lambert-W-function does not help here. It is very likely that you can only approximate the value. The newton method quickly leads to a good approximation. – Peter Jan 10 '18 at 10:11
  • An interesting question is whether the solution is transcendental. – Peter Jan 10 '18 at 10:13

1 Answers1

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If $(2, x+1)=1$

also $(3, x+1)=1$ we may write:

$2^x + 3^x ≡ 2 mod (x+1)$ or $2^x+3^x= y(x+1) +2=10000 $

⇒ $y(x+1)=9998$

Now we have a system of equation:

$y(x+1)=9998$

$2^x+3^x=10000$

That is the solution is the coordinates of intersection of two curves; one a hyperbola, another an exponent curve, hence the equation can be solved by sketching.

Also:

$2^8+3^8=6817$

$2^9+3^9=20195$

⇒ $8 < x < 9$

by try and error we can find that $x≈ 8.3535$

$2^{8.3535}+3^{8.3535}≈ 10001.66$

Hence $y=\frac{9998}{9.3535}≈1069$

sirous
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