Question

Write the solution set of the following inequlity :

$\displaystyle{\dfrac{1}{2^x-1}\ \textgreater \ \frac{1}{1-2^{x-1}}}$

Answer 1

Solution:

The given inequality is

    $\frac{1}{2^{x}-1}>\frac{1}{1-2^{x-1}}$

If we take x < 0, we will face a contradiction.

Let, x = - 1

L.H.S. = $\frac{1}{2^{-1}-1}$ = - 2 &

R.H.S. = $\frac{1}{1-2^{-1-1}}$ = 4/3

But - 2 is not greater than 4/3

So we take x > 0

Now, $\frac{1}{2^{x}-1}>\frac{1}{1-2^{x-1}}$

   or, $\frac{1}{2^{x}-1}>-\frac{2}{2^{x}-2}$

   or, $\frac{1}{2^{x}-1}+\frac{2}{2^{x}-2}>0$

   or, $\frac{2^{x}-2+2.2^{x}-2}{(2^{x}-1)(2^{x}-2)}>0$

   or, $\frac{3.2^{x}-4}{(2^{x}-1)(2^{x}-2)}>0$

Since x > 0, we take both the numerator and the denominator to be > 0, because if we take both of them to be < 0, we will attain a value from $2^{x}-1 <0$ that x < 0, a contradiction.

Now, $3.2^{x}-4>0$

   or, $2^{x}>\frac{4}{3}$

   or, $x>\frac{log\frac{4}{3}}{log2}$ ,

$2^{x}-2>0$

or, $2^{x}>2^{1}$

or, x > 1 and

$2^{x}-1>0$

or, $2^{x}>2^{0}$

or, x > 0

Therefore, the required solution is

    0 < x < $\frac{log\frac{4}{3}}{log2}$ , x > 1