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?
On
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.
On
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).
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.