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Solve the following inequality

$$(3^x-2^x)^2+3^x-2^x\leq 5^x$$

I'm having difficulty in solving this inequality because there are multiple exponential with different basis. I thought, however, the replacement:

$$y=3^x-2^x$$

and so:

$$y^2+y\leq 5^x$$

treating $5^x$ as the known term of the inequality of the second degree in y.

What do you think? thank you very much

Mark
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2 Answers2

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Your approach will flesh out as follows:

$$\boxed{y^2+y\le5^x}\iff y^2+y-5^x\le0\iff y\in\Big[y_{_1},y_{_2}\Big],\qquad y_{_{1,2}}=\frac{-1\pm\sqrt{5^x+1}}2$$

$$-\sqrt{5^x+1}\le2y+1\le\sqrt{5^x+1}\iff|2y+1|\le|\sqrt{5^x+1}|\iff(2y+1)^2\le5^x+1$$

$$4y^2+4y\le5^x\iff\boxed{y^2+y\le\dfrac{5^x}{4\ \ }}\iff y\in\Big[y'_{_1},y'_{_2}\Big],\qquad y'_{_{1,2}}=\frac{-1\pm\sqrt{\dfrac{5^x}{4\ \ }+1}}2$$

$$4y^2+4y\le\dfrac{5^x}{4\ \ }\iff\boxed{y^2+y\le\dfrac{5^x}{4^2}}\iff y\in\Big[y^"_{_1},y^"_{_2}\Big],\qquad y^"_{_{1,2}}=\frac{-1\pm\sqrt{\frac{5^x}{4^2}+1}}2$$

Traveling down this road, we deduce that $y\in\Big[Y_{_1},Y_{_2}\Big]$, where $Y_{_{1,2}}=\displaystyle\lim_{n\to\infty}\dfrac{-1\pm\sqrt{\frac{5^x}{4^n}+1}}2=\pm1.$

So $-1\le3^x-2^x\le1\iff2^x-1\le3^x\le2^x+1$. Since $\displaystyle\lim_{t\to-\infty}a^t=0\iff x\in(-\infty,x_{_0})$, for some numerically determined value of $x_{_0}$.

Lucian
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  • Excuse me, but I did not understand the last step. Also I do not understand what is the range where the inequality is verified. – Mark Feb 02 '14 at 20:41
  • @Mark: My point was that your approach is as correct as it is useless, since it is ultimately equivalent to saying $y^2+y\le0$, which is self-evident, since we already know that $y^2+y\le5^x$, and $5^x>0$ for all $x\in\mathbb{R}$. – Lucian Feb 02 '14 at 23:11
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By inspection, for $x=0$ as well as for $x=1$, the inequality is satisfied but it is not for $x=2$. So, what it looks is that this inequality is satisfied for any value of $x$ which is lower than the solution of $$(3^x-2^x)^2+3^x-2^x= 5^x$$ The problem is that this solution cannot be obtained analytically.

To obtain the solution of this equation, you can use Newton method starting at $x=2$; the successive iterates are : $2.00000$, $1.86852$, $1.82445$, $1.82013$, $1.82009$.

Let me know if you need me to elaborate in more details about Newton iterative method for solving equations.

Claude Leibovici
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