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Solve the equation $2^x - 3^{x-1}=-(x+2)^2$

How I got this question? I created this question so I know the answer. The answer is 5. But I have no idea how to solve it. Take note that I cannot do logarithm, guess and check and modulus. Does anybody know how to solve this? I have no idea how to start.

$2^x - 3^{x-1}$ is infactorisable. Even if I did $2^x - (2+2^0)^{x-1}$, it is STILL infactorisable. I was hoping to solve by hand.

So, I do I solve the equation?

ministic2001
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  • it's not doable analyitically.. you're just lucky it's an integer but here you should draw the graphics of the two functions and guess a numerical range for the solution – Nick Feb 20 '16 at 12:16
  • @NicolòRuggeri so you are basically saying you can only solve this by graph? – ministic2001 Feb 20 '16 at 12:17
  • I've never seen it done another way.. it's like solving $e^x=x$ and all the other equations you can't easily solve by isolating x.. This is why in all the calculus courses the give you the zero existence theorem and stuff like that – Nick Feb 20 '16 at 12:19
  • But there could be an original way for this particular case anyway – Nick Feb 20 '16 at 12:20
  • Equation like these cannot be solved explicitly (unless you can use Lambert's W and other special function) so you usually "solve it graphically" but in this case there is a simple answer $5$ so I think is just a matter of guessing and trying different integers solution. – AlienRem Feb 20 '16 at 12:45
  • I saw a split second someone using log to solve it. I would want to see how you solve, but I'll still tick SS_C4 if no one solved by hand without log and modulus and without guessing – ministic2001 Feb 20 '16 at 12:46

3 Answers3

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Well, from $2^x - 3^{x-1} = -(x+2)^2$, $2^x = 3^{x-1} - (x+2)^2$.

LHS is always even, and $3^n$ is always odd. Therefore $(x+2)^2$ has to be odd, $\Rightarrow$ x is odd.

Also, from the first equation, $2^x < 3^{x-1}$. This is true for $x > 2$. Since x is odd, the new condition for $x$ is $x \ge 3$ and x is odd. So, $$x \in {\{3,5,7,9,\dots\}}$$

Checking, we see that $5$ is an answer.

(But I'm not sure how to prove that this is the only answer. Maybe I could bring the graph part here).

SS_C4
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HINT.-You have $$3^{x-1}-(x+2)^2=2^x$$ Searching for an eventual rational solution one has $$(3^{\frac{x-1}{2}}+x+2)(3^{\frac{x-1}{2}}-x-2)=2^x$$ We do now $$3^{\frac{x-1}{2}}+x+2=2^{x-h}$$ $$3^{\frac{x-1}{2}}-x-2=2^h$$ Hence $$2x+4=2^{x-h}-2^h\iff x+2=2^{x-h-1}-2^{h-1}\qquad (*)$$ Trying with $h=1$ we find out the easy $$x+2=2^{x-2}-1$$ where one can see the integer solution $x=5$.

For values $h\gt 1$ we have in $(*)$ that $x$ is necessarily even excepting when $x=h+1$. In the first case this contradicts the original equation. I stop here. I wanted just to show the value $x=5$ can be deduced.

Piquito
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The only solution is $x=5$.

You can see this by graphing the LHS and RHS separately and seeing that there is only one point of intersection. Graph

SS_C4
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  • So there is no way to do by hand.. (Im aware you used "Desmos" app) I'm hoping for other answers to see how tey do by hand. If there are no other possible way, I will tick yours as complete answer. – ministic2001 Feb 20 '16 at 12:40
  • Alright, but I'll try to find another solution myself. – SS_C4 Feb 20 '16 at 12:41