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$2 \cdot 3^x + 2 = 5 \cdot 2^x$ I am looking for the exact real solutions of this equation(no Newton's method). Even my professor couldn't solve this. I myself have struggled a lot to find the solutions. Can anyone give me any ideas? (This is my first question on this forum so I'm sorry for the formatting)

CatPan26
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    $x = 2$ is a solution obtained by inspection. – user0102 Feb 10 '21 at 06:39
  • @PTDS I already have. I am looking for a solution that doesn't 100% rely on the computed plot. – CatPan26 Feb 10 '21 at 06:47
  • Do you want integer solutions? – jojobo Feb 10 '21 at 07:01
  • @jojobo I am looking for real solutions, not only integers – CatPan26 Feb 10 '21 at 07:05
  • One can show using calculus that $2\cdot3^x+2-5\cdot2^x$ has a unique local minimum that is negative (you can even find it exactly) and is decreasing to the left of that minimum and increasing to its right; the function also has limit $2$ as $x\to-\infty$ and $\infty$ as $x\to\infty$. Therefore it has exactly two real roots. The negative one, near $-0.8$, doesn't seem to have any nice form, but of course Newton's method will yield as many decimal places as we want. – Greg Martin Feb 10 '21 at 07:11
  • I haveto find the exact solutions, so Newton's method doesn't seem to be an option – CatPan26 Feb 10 '21 at 07:18
  • Using the "change-of-base" relations, we could write this equation as $$ 5 · 2^x \ - \ 2 · 2^{(\log_2 3) · x } \ \ = \ \ 2 \ \ . $$ But that transcendental number in the exponent in the second term makes me pretty dubious that you can write a tidy exact second root. –  Feb 10 '21 at 07:30

1 Answers1

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Excluding the trivial solution,inspection shows that function $$y=2 \cdot 3^x + 2 - 5 \cdot 2^x$$ has a root "close" to $x=-1$.

Build a series expansion around this point to have $$y=\frac 16+\sum_{n=1}^p \frac{a_n}{n!}\,(x+1)^n+O\left((x+1)^{p+1}\right)$$ where $$a_n=\frac 2 3 \big[\log(3)\big]^n-\frac 52 \big[\log(5)\big]^n$$ Now, make a series reversion to obtain $$x=-1+\sum_{n=1}^p b_n \left(\frac{y-\frac{1}{6}}{\frac{2 \log (3)}{3}-\frac{5 \log (2)}{2}}\right)^n+O\left(\left(y-\frac{1}{6}\right)^{p+1}\right)$$ Now, make $y=0$ and you have long explicit forms of more and more accurate solutions.

Here are the results of this process $$\left( \begin{array}{cc} p & x_{(p)} \\ 1 & -0.8334099246 \\ 2 & -0.8389092790 \\ 3 & -0.8385065961 \\ 4 & -0.8385304569 \\ 5 & -0.8385285436 \\ 6 & -0.8385286777 \\ 7 & -0.8385286666 \\ 8 & -0.8385286675 \\ 9 & -0.8385286674 \end{array} \right)$$

Edit

As I replied to a comment, there is no closed form. But the exact solution is obtained using summations up to infinity.

Claude Leibovici
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  • But OP said he was looking for an exact solution. We actually know that there is no closed form, but we cannot prove it. – lone student Feb 10 '21 at 08:49
  • @lonestudent. There is no closed form : this is partly OK. But, there is an exact solution if you make my summations up to infinity. – Claude Leibovici Feb 10 '21 at 08:52
  • Well thank you! I suppose there isn't an exact solution after all:)) – CatPan26 Feb 10 '21 at 08:54
  • @CatPan26. NO, it is ! Look at my edit. Cheers :-) – Claude Leibovici Feb 10 '21 at 08:56
  • Yes, there is definitely no closed form. Let me state my question more precisely. For example, "There is no closed form that use the function Lambert W".Can we prove this claim? What tools do we have to prove this?I upvoted your answer. But for me the striking answer would be proof that the closed form cannot exist. For example, the integral $\int \dfrac {\sin x}{x}$ can not be solved elementary function. And it is possible to prove this with Galois theory.I am trying to explain what I want to ask by making metaphors. $x^5-x+1$ cannot be solved by radicals. We have a proof for this isn't it? – lone student Feb 10 '21 at 09:11
  • @lonestudent. I think that you want to start a discussion which could last a few years (or more). So, be gentle since I am 80 !! Back to serious, do you consider that an hypergeometric function would be an acceptable solution ? I must confess that the concept of non-elementary functions ran out of my mind almost 60 years ago (I am not a teacher). – Claude Leibovici Feb 10 '21 at 09:16
  • This is definitely the result of my bad English. I am definitely someone who always bows to my teachers with respect. I'm looking again with google translate. I check my comment if there is a word that is rude/wrong in the comment. By the way, I know that the information you have is at academic level. I only have school math. I have nothing else. I cannot stop asking questions just out of curiosity. I wish you well-being, health and well-being. Thanks for comments. My grandmother is now 80 years old too. God bless you. – lone student Feb 10 '21 at 09:25
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    @lonestudent. There is no problem for me. I shall not speak about my terrible English (by the way, if I may ask, where are you located ? I am in Pau (SO of France, close to Spain). Feel free to ask me as many questions as you wish. If you want, make a list of items and, one of these days, open a chat room and ping me. It will be my pleasure. Cheers :-) – Claude Leibovici Feb 10 '21 at 09:33