Two logarithmic functions are added in this logarithmic problem and it is given that their sum is equal to the quotient of $5$ by $2$.

$\log_{5}{(x)}+\log_{x}{(5)}$ $\,=\,$ $\dfrac{5}{2}$

The log equation is expressed in terms of a variable $x$. so, the value of indeterminate $x$ should be calculated by solving the given logarithmic equation.

In the logarithmic expression, each term consists of the variable $x$ and the number $5$ but the places of two quantities in log terms are opposite. Hence, the logarithmic terms cannot be added directly due to the unlikeness of them.

The unlikeness of the terms can be removed by changing the base of any one of the log terms as per the switch rule of logarithms.

$\implies$ $\log_{5}{x}$ $+$ $\dfrac{1}{\log_{5}{x}}$ $\,=\,$ $\dfrac{5}{2}$

Now, the two log terms in the equation contain same number and same base but they cannot be added directly due to their reciprocity.

Assume, $y = \log_{5}{x}$ for avoiding the confusion in simplifying the log equation.

Now, replace each log term by a variable $y$ to express the logarithmic equation in simple form.

$\implies$ $y+\dfrac{1}{y}$ $\,=\,$ $\dfrac{5}{2}$

$\implies$ $\dfrac{y^2+1}{y}$ $\,=\,$ $\dfrac{5}{2}$

$\implies$ $2(y^2+1) = 5y$

$\implies$ $2y^2+2-5y=0$

$\,\,\,\therefore\,\,\,\,\,\,$ $2y^2-5y+2=0$

The given logarithmic equation in terms of $x$ is successfully converted as a quadratic equation in terms of $y$.

$2y^2-5y+2=0$

This quadratic equation can be solved by the factoring method.

$\implies$ $2y^2-4y-y+2$ $\,=\,$ $0$

$\implies$ $2y(y-2)-y+2$ $\,=\,$ $0$

$\implies$ $2y(y-2)-(y-2)$ $\,=\,$ $0$

$\implies$ $2y(y-2)-1(y-2)$ $\,=\,$ $0$

$\implies$ $(y-2)(2y-1)$ $\,=\,$ $0$

$\implies$ $y-2 \,=\, 0$ or $2y-1 \,=\, 0$

$\implies$ $y \,=\, 2$ or $2y \,=\, 1$

$\,\,\,\therefore\,\,\,$ $y \,=\, 2$ or $y \,=\, \dfrac{1}{2}$

Therefore, the solution set for the variable $y \,=\, \bigg\{\dfrac{1}{2}, 2\bigg\}$

In fact, the equation in this problem is given in terms of $x$ but we have calculated the solution in terms of $y$. Therefore, it is time to obtain the solution in terms of $x$.

Now, substitute the value of $y$ in terms of $x$ in the solution.

$\implies$ $\log_{5}{x} \,=\, 2$ or $\log_{5}{x} \,=\, \dfrac{1}{2}$

For solving the variable $x$, express each logarithmic equation into an exponential equation as per the mathematical relation between the logarithms and exponents.

$\implies$ $x \,=\, 5^2$ or $x \,=\, 5^{\Large \frac{1}{2}}$

$\,\,\,\therefore\,\,\,$ $x \,=\, 25$ or $x \,=\, \sqrt{5}$

Therefore, the solution set for the variable $x \,=\, \Big\{\sqrt{5}, 25\Big\}$

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