Solve the inequality $\frac {(\frac 2 3)^{x-1}-1}{\sqrt2-\sqrt[3]{2^{x-1}}} < 0$ 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?
 A: 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 $1<x<\frac52$.
A: 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?
