How do you evaluate ${{\log }_{\dfrac{1}{5}}}25?$

Hint: Logarithm function is defined as the inverse of function of exponentiation. Consider again the exponential function $f(x)={{b}^{4}}$ Where $b>0Log bases of $b$ of $x.$ logarithmic is useful because they enable us to perform calculations with very large numbers.

Complete step-by-step answer:
${{\log }_{\dfrac{1}{5}}}25$
Rewrite this as an equation
${{\log }_{\dfrac{1}{5}}}25=x$
Now Rewrite ${{\log }_{\dfrac{1}{5}}}\left( 25 \right)=x$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real number and $b$ does not equal $1$ then ${{\log }_{b}}\left( n \right)=y$ is equivalent to ${{b}^{4}}=n$
$\dfrac{1}{5x}={{5}^{2}}$ or ${{\left( \dfrac{1}{5} \right)}^{x}}={{5}^{2}}$
${{\left( {{5}^{-1}} \right)}^{x}}={{5}^{2}}$
Create an equivalent expression in the equation that all have an equal base. Rewrite ${{\left( {{5}^{-1}} \right)}^{n}}$ as.
${{5}^{-x}}={{5}^{2}}$
Since the bases are the same the two expressions are only equal if the exponents are also equal.
$-x=2$
$x=-2$
The variable $n$ is equal to $-2$

Hence ${{\log }_{\dfrac{1}{5}}}25=-2$

Additional Information:
We can solve any Numerical of this type with this exponent logarithm. We will solve one more numerical with a different base with the same method.
Example: ${{\log }_{25}}\left( \dfrac{1}{5} \right)$
Rewrite this equation.
${{\log }_{25}}\left( \dfrac{1}{25} \right)=x$
Rewrite ${{\log }_{25}}\left( \dfrac{1}{25} \right)=x$ in exponential form using the definition of a logarithm if $n$ and $b$ are positive real number and $b$ does not equal $1$ then ${{\log }_{b}}\left( x \right)=y$ equivalent to ${{b}^{-4}}=x$
${{b}^{4}}=x$
${{25}^{n}}=\dfrac{1}{5}$
Create expression in the equation that all have equal bases ${{\left( {{5}^{2}} \right)}^{n}}={{5}^{-1}}$
Rewrite ${{\left( {{5}^{2}} \right)}^{x}}$ as ${{5}^{2x}}$
${{5}^{2x}}={{5}^{-1}}$
Since the bases are the same then two expressions are only equal if the exponents are also equal.
$2x=-1$
Solve this for $n$
$n=-\dfrac{1}{2}$
The variable is $x$ is equal to $-\dfrac{1}{2}$
The result can be shown in multiple forms exact form $x=-\dfrac{1}{2}$ and Decimal form $x=-0.5$

Note:
Students make mistakes with logarithms because they are working with exponents in reverse. This is challenging four our brains since we often are not so confident with our powers of numbers and the exponent properties. Now power of $10$ is easy for us right now. Just count the number of zeros to the right of the $1$ for the positive exponent and move the decimal to the left for negative exponents. Therefore a student who knows power of $10$ just as well $\log \left( 10 \right)=1$ which is same as ${{\log }_{\left( 10 \right)}}\left( 10 \right)=1$
But let’s try some other bases ${{2}^{3}}=8$ so ${{\log }_{2}}\left( 8 \right)=3$ since the answer to the logarithms is the power of $2$ that equal $8.$ The answer to $\log $ is the exponent keeps in mind.