Solving an exponential equation with absolute value

Question

I am to solve this equation: $|\frac{-2^x}{1-2^x}| < 1.$
And so I got rid of the modus sign: $\frac{-2^x}{1-2^x} < 1 $
or $\frac{-2^x}{1-2^x} > -1$

But I am stuck now. How should I continue?

Answer 1

You have $$-1\lt\frac{-2^x}{1-2^x}\lt 1\tag1$$ If $1-2^x\lt 0$, i.e. $x\gt 0$, $$(1)\iff -1+2^x\gt -2^x\gt 1-2^x$$ There is no such $x$.

If $1-2^x\gt 0$, i.e. $x\lt 0$, $$(1)\iff -1+2^x\lt -2^x\lt 1-2^x\iff x\lt -1$$

Hence, the answer is $x\lt -1$.

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Another way :

$$\begin{align}\left|\frac{-2^x}{1-2^x}\right|\lt 1&\iff |-2^x|\lt |1-2^x|\quad\text{and}\quad 1-2^x\not=0\\&\iff (-2^x)^2\lt (1-2^x)^2\quad\text{and}\quad x\not=0\\&\iff 2^{x+1}\lt 1\quad\text{and}\quad x\not=0\\&\iff x\lt -1\end{align}$$

Answer 2

Hint put $2^x=y$ and then get appropriate range.

Answer 3

$$\left|\dfrac{-2^x}{1-2^x}\right|=\left|\dfrac{2^x}{1-2^x}\right|=\dfrac1{|2^{-x}-1|}$$

We need $|2^{-x}-1|>1$

$\implies$ either $2^{-x}-1>1\iff2^{-x}>2^1\iff -x>1\iff x<-1$

or $2^{x-1}-1<-1\iff2^{-x}<0$ which is impossible for real $x$ as $2^{-x}>0$