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I found an interesting equation I don't seem to grasp completely.

Solve the equation:

$2^x+3^x-4^x+6^x-9^x=1$

Now. I thought the only relevant solution was the integer solutions, but I was sorely mistaken.

So if the question was: Find all integer solutions to the following equation my attempt is as follows:

$2^x+3^x-4^x+6^x-9^x=1$ (1)

$2^x+3^x+3^x2^x-(2^x)^2-(3^x)^2=1$

Substitution:

$P=2^x\quad$ $Q=3^x$ (2)

$P+Q+PQ-P^2-Q^2=1$

We can see that this equation is symmetric and that $P>0, Q>0$

For $P=1$ we get $1+Q+Q-1-Q^2=1 \to Q=1$

P & Q are equal, with (2) we get: $P=Q \to 2^x=3^x \to x=0$

Answer: $x=0$ is the only solution

Now when I wrote down my solution I recognize that several steps are wrong. First of all how do i know P and Q are equal? They might not be i suppose

If anyone could give me a hint on how to solve these type of equations or even how to solve this equation I'd be very grateful.

And furthermore is this type of equation called exponential equation?

The type of exponential equation I'm more familiar with is simply solved with log rules, equations were you can simplify both sides and just log and solve.

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  • $\begingroup$ Note that, if $x> 0$ then $P$ is even whence $\mod 2$ your expression is $ Q-Q^2\equiv 1 \pmod 2$ which is impossible. Were you restricting to non-negative $x$? $\endgroup$ – lulu Jun 19 '17 at 16:42
  • $\begingroup$ @lulu $P$ and $Q$ may not be integers. $\endgroup$ – Bob Krueger Jun 19 '17 at 16:45
  • $\begingroup$ I was, but for the complete solution we must work with all reels i think. $\endgroup$ – einar Jun 19 '17 at 16:47
  • $\begingroup$ @Bob1123 They are if $x\in \mathbb N$ that's why I asked if we restricting to non-negative integers $x$. $\endgroup$ – lulu Jun 19 '17 at 16:48
  • $\begingroup$ @einar But you still want $x$ to be an integer, no? I mean, you write "Find all integer solutions to the following..." If $x=-n$ for $n\in \mathbb N$ then you can multiply both sides by $6^{2n}$ to get $2^n3^{2n}+2^{2n}3^n-3^{2n}+6^n-2^{2n}=6^{2n}$ which again is impossible $\pmod 2$. $\endgroup$ – lulu Jun 19 '17 at 16:51
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Note that the equation is equivalent to $$(2^x-1)^2-(2^x-1)(3^x-1)+(3^x-1)^2=0.$$ Moreover, note that the above equation is satisfied iff $x=0$ because if $v\not=0$ then $$u^2-uv+v^2=v^2((u/v)^2-(u/v)+1)>0.$$

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  • $\begingroup$ By subtracting $(2^x-1)(3^x-1)$, we get a perfect square on the LHS, and something that is $\le0$ on the RHS, therefore, $0=(2^x-1)(3^x-1)$, and thus it follows suit. Beautiful! $\endgroup$ – Simply Beautiful Art Jun 19 '17 at 16:50
  • $\begingroup$ Thanks! That's a beautiful solution indeed! :) Much appreciated! $\endgroup$ – einar Jun 19 '17 at 16:53
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    $\begingroup$ @SimplyBeautifulArt this is what happens when you write the conic section in $P,Q$ around its geometric center, located where the gradient is zero $\endgroup$ – Will Jagy Jun 19 '17 at 16:54
  • $\begingroup$ @SimplyBeautifulArt Yeah it's very nice! You can also view it as $\dfrac {a^3+b^3}{a+b}=0$ from which you get $2^x + 3^x=2$, which has $x=0$ as the only solution. $\endgroup$ – Ovi Jun 19 '17 at 16:56
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    $\begingroup$ @WillJagy Ah thanks a lot for explaining the motivation for this. $\endgroup$ – Ovi Jun 19 '17 at 16:57
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Given conic section $$ P^2 - PQ + P^2 - P - Q + 1 = 0, $$ the $P$ partial derivative is $$ 2P-Q - 1. $$ The $Q$ partial is $$ -P + 2Q - 1. $$ Setting both to zero gives $P=Q=1.$

So $$ P = u + 1, \; \; Q = v + 1. $$

$$ P^2 - PQ + P^2 - P - Q + 1 = u^2 - uv + v^2. $$ The set $u^2 - uv + v^2 = 0$ is a single point $u=v=0,$ as this is a positive quadratic form: the Hessian matrix is positive.

Oh, if the Hessian has determinant zero, there will not be a "center," the thing is a parabola or degenerate parabola.

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