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I'd like to solve the following inequality:

$$\frac {(\frac 2 3)^{x-1}-1}{\sqrt2-\sqrt[3]{2^{x-1}}} < 0$$

I made it so that

$$z = 2^{x-1}$$

This is what the inequality now looks like:

$$\frac {\frac z {3^{x-1}}}{2^{\frac 1 2}-z^{\frac 1 3}} < 0$$

I'm stuck. Any hints on what's the best approach to solve this inequality?

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It's best to break this apart by looking at the numerator and denominator separately, since we already have $0$ on one side of the inequality. Let's look at the numerator first. We'll determine when it's equal to $0.$ \begin{align*} \left(\frac{2}{3}\right)^{x-1} - 1 &= 0\\[0.3cm] \left(\frac{2}{3}\right)^{x-1} &= 1\\[0.3cm] x-1 &= 0\\ x &= 1 \end{align*}

Now let's see when the denominator equals $0.$ \begin{align*} \sqrt{2} - \sqrt[3]{2^{x-1}} &= 0\\[0.3cm] 2^{1/2} &= 2^{(x-1)/3}\\[0.3cm] \left(2^{1/2}\right)^6 &= \left(2^{(x-1)/3}\right)^6\\[0.3cm] 8 &= 2^{2x-2}\\ 2x-2 &= 3\\ x &= 5/2 \end{align*}

So the numerator changes sign at $x = 1$ and the denominator changes sign at $x = 5/2$. Can you take it from here?

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  • $\begingroup$ Excellent explanation, thanks! $\endgroup$ – Cesare Jul 12 '16 at 15:03
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For the fractions to be negative you want the numerator and denominator to have opposite signs. Consider each separately is easier.

$\left(\frac23\right)^{x-1}-1$ will change sign at $x=1$ as $\left(\frac23\right)^0=1$. Checking values either side shows for $x<1$ the numerator is positive and for $x>1$ the numerator is negative.

$\sqrt2-\sqrt[3]{2^{x-1}}$ will change sign at $x=\frac52$ as $\sqrt[3]{2^{\frac52-1}}=\sqrt2$. Checking values either side shows for $x<\frac52$ the numerator is positive and for $x>\frac52$ the numerator is negative.

So look for when they have opposite signs.

Hence the fraction is negative when $1<x<\frac52$.

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