The set of all values of $x$ satisfying the inequality $\log _{\left(x+\frac{1}{x}\right)}\left[\log _{2}\left(\frac{x-1}{x+2}\right)\right]>0$ is

For $\log_b(a)$ to exist, we need $b > 0$, $b \neq 1$, and $a > 0$.

For $\log_b(a) > 0$, there are two cases:

• If $b > 1$, then $a > 1$
• If $0 < b < 1$, then $0 < a < 1$

Let us label the parts:

Outer base: $B = x + \dfrac{1}{x}$

Outer argument: $A = \log_2\left(\dfrac{x-1}{x+2}\right)$

Our inequality is $\log_B(A) > 0$.

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For $\log_2\left(\dfrac{x-1}{x+2}\right)$ to exist, we need:

$\dfrac{x-1}{x+2} > 0$

Using a sign chart:

Interval	$x - 1$	$x + 2$	Fraction
$x < -2$	$-$	$-$	$+$
$-2 < x < 1$	$-$	$+$	$-$
$x > 1$	$+$	$+$	$+$

So $x \in (-\infty, -2) \cup (1, \infty) \quad \cdots (i)$

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For the outer argument $A > 0$, we need:

$\log_2\left(\dfrac{x-1}{x+2}\right) > 0$

Since base $2 > 1$, this gives us:

$\dfrac{x-1}{x+2} > 1$

$\dfrac{x-1}{x+2} - 1 > 0$

$\dfrac{x - 1 - x - 2}{x+2} > 0$

$\dfrac{-3}{x+2} > 0$

Since $-3 < 0$, we need $x + 2 < 0$, giving us $x < -2$.

So $A > 0$ requires $x \in (-\infty, -2) \quad \cdots (ii)$

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Now we check the outer base $B = x + \dfrac{1}{x}$ on the interval $x < -2$.

When $x < -2$, both $x$ and $\dfrac{1}{x}$ are negative, so:

$x + \dfrac{1}{x} < 0 \quad \text{for all } x < -2$

A logarithm's base must be positive, so the outer logarithm is undefined for all $x < -2$.

This means whenever $A > 0$ (which needs $x < -2$), the base $B$ is negative and the expression does not exist.

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Let us check the remaining interval $x > 1$ from condition $(i)$.

Here $B = x + \dfrac{1}{x} > 0$ since $x > 1$, so the base is valid.

However, for $x > 1$:

$\dfrac{x-1}{x+2} = \dfrac{\text{positive}}{\text{positive}}$, but $x - 1 < x + 2$, so $\dfrac{x-1}{x+2} < 1$.

This gives $\log_2\left(\dfrac{x-1}{x+2}\right) < 0$, meaning $A < 0$.

Since the argument of the outer logarithm is negative, $\log_B(A)$ is undefined for $x > 1$ as well.

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Summarizing:

Interval from domain	Is $A > 0$?	Is $B > 0$?	Conclusion
$x < -2$	Yes	No $(B < 0)$	Outer log undefined
$x > 1$	No $(A < 0)$	Yes	Outer log undefined

The two conditions needed simultaneously — a positive base and a positive argument for the outer logarithm — are satisfied on completely different intervals with no overlap.

$x \in \emptyset$