Solving the Logarithm Inequality \(x^{{{\log }_{7}}x}} > 7\)

We are asked to find the values of \(x > 0\) that satisfy the inequality \({{x}^{{{\log }_{7}}x}}>7\).

The given inequality is:

\(\qquad {{x}^{{{\log }_{7}}x}}>7\)

Since both sides are positive (as \(x > 0\)), we can take the logarithm of both sides. Using the base-7 logarithm is convenient because of the term \({{\log }_{7}}x}\) in the exponent. The base 7 is greater than 1, so taking the base-7 logarithm will preserve the direction of the inequality.

Taking \({{\log }_{7}}\) on both sides:

\(\qquad {{\log }_{7}}({{x}^{{{\log }_{7}}x}}})>{{\log }_{7}}(7)\)

Using the logarithm property \({{\log }_{b}}({{a}^{c}})=c\cdot {{\log }_{b}}(a)\), we can bring the exponent \({{\log }_{7}}x}\) down:

\(\qquad ({{\log }_{7}}x}) \cdot ({{\log }_{7}}x}) > 1\)

This simplifies to:

\(\qquad ({{{\log }_{7}}x})^2 > 1\)

Let \(y = {{\log }_{7}}x}\). The inequality becomes:

\(\qquad {{y}^{2}}>1\)

To solve this quadratic inequality, we can rearrange it:

\(\qquad {{y}^{2}}-1 > 0\)

Factor the left side as a difference of squares:

\(\qquad (y-1)(y+1) > 0\)

For the product of two factors to be positive, both factors must be positive or both must be negative.

• Case 1: Both factors are positive.

  \(y-1 > 0\) and \(y+1 > 0\)

  \(y > 1\) and \(y > -1\)

  Both conditions are satisfied when \(y > 1\).

• Case 2: Both factors are negative.

  \(y-1 < 0\) and \(y+1 < 0\)

  \(y < 1\) and \(y < -1\)

  Both conditions are satisfied when \(y < -1\).

So, the solution for \(y\) is \(y < -1\) or \(y > 1\).

Now, substitute back \(y = {{\log }_{7}}x}\):

\(\qquad {{\log }_{7}}x < -1 \quad \text{or} \quad {{\log }_{7}}x > 1\)

Since the base of the logarithm is 7, which is greater than 1, the logarithmic function \({{\log }_{7}}x}\) is an increasing function. We can convert these logarithmic inequalities back to inequalities involving \(x\) by raising 7 to the power of both sides, maintaining the inequality direction.

For \({{\log }_{7}}x < -1\):

\(\qquad x < {{7}^{-1}}\)

\(\qquad x < \frac{1}{7}\)

Combining this with the initial condition \(x > 0\), we get \(0 < x < \frac{1}{7}\). This interval is written as \(\left(0, \frac{1}{7}\right)\).

For \({{\log }_{7}}x > 1\):

\(\qquad x > {{7}^{1}}\)

\(\qquad x > 7\)

This interval is written as \(\left(7, \infty\right)\).

Combining the two possible intervals for \(x\), the solution is the union of these intervals:

\(\qquad x \in \left(0, \frac{1}{7}\right) \cup \left(7, \infty\right)\)

This corresponds to option 3.