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**Hint:**If a>$1$ ,

then ${{\log }_{a}}x>{{\log }_{a}}y$

$\Rightarrow $ x > y

and if $0$ < a < $1$

then ${{\log }_{a}}x>{{\log }_{a}}y$

$\Rightarrow $ x < y.

This means that when we take antilog on both sides of the equation, we have to reverse the inequalities. Also note that

${{\log }_{a}}{{a}^{k}}=k$ ……($2$)

and

${{a}^{(\log {{a}^{k}})}}=k$ ……($3$)

for any number k and a > $0$

**Complete step by step solution:**

We first raise both sides to the power of $\dfrac{1}{4}$. Then using the Hint we get

$\begin{align}

& {{\dfrac{1}{4}}^{{{\log }_{\dfrac{1}{4}}}(\dfrac{35-{{x}^{2}}}{x})}}\le {{\dfrac{1}{4}}^{-\dfrac{1}{2}}} \\

& \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le {{\dfrac{1}{4}}^{-\dfrac{1}{2}}} \\

& \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le {{4}^{\dfrac{1}{2}}} \\

& \Rightarrow (\dfrac{35-{{x}^{2}}}{x})\le 2 \\

\end{align}$

with the inequality reversed as $0<\dfrac{1}{4}<1$. We have also used the fact that

$\dfrac{1}{{{a}^{k}}}={{a}^{-k}}$ for any numbers $a$ and $k$.

Now the inequality is a simple polynomial inequality and can be solved as

$\begin{align}

& (\dfrac{35-{{x}^{2}}}{x})\le 2 \\

& \Rightarrow 35-{{x}^{2}}\le 2x \\

& \Rightarrow {{x}^{2}}+2x-35\ge 0 \\

\end{align}$

This quadratic equation can be easily factorized which gives

$\begin{align}

& {{x}^{2}}+(7-5)x-(7\times 5)\ge 0 \\

& \Rightarrow (x-5)(x+7)\ge 0 \\

\end{align}$

In the last step we only need to realize that this expression is positive only when either of the terms are positive or when both are negative. Since x > $5$ clearly means that x > $-7$ we have one range of solutions as x >$5$ . Also x < $-7$ would also make the first factor negative so we have the other range as x < $-7$. The middle part has the expression negative since the first bracket is negative and second positive.

So we write the solutions as

$x\in (-\infty ,-7]\cup [5,\infty )$

**Note:**

The reversal of inequality is necessary. A common error would be to not do that and that would lead to the erroneous solution of x between $-7$ and $5$.

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