Question

How do you evaluate \[{\log _{\dfrac{2}{5}}}\left( {\dfrac{{125}}{8}} \right)\]?

Answer 1

Hint: First we will convert this equation into the form ${\log _a}b$. Then we will evaluate all the required terms. Then we will apply the property. Here, we are using ${({\log _a}{b^{ - 1}})^m} = {({\log _a}b)^{ - m}}$ and ${\log _a}{b^m} = mlo{g_a}b$ logarithmic property. The value of the logarithmic function $\ln e$ is $1$.

Complete step by step answer:
We will first apply the logarithmic property to convert the equation to solvable form. Compare the given equation with formula and evaluate the values of the terms.
\[ = {\log _{\dfrac{2}{5}}}\left( {\dfrac{{125}}{8}} \right)\]
Hence, here the values are:
$
\Rightarrow a = \dfrac{2}{5} \\
\Rightarrow b = \dfrac{{125}}{8} \\
 $
Now we will try to simplify the given term.
\[
\Rightarrow {\log _{\dfrac{2}{5}}}\left( {\dfrac{{125}}{8}} \right) \\
\Rightarrow {\log _{\dfrac{2}{5}}}{\left( {\dfrac{5}{2}} \right)^3} \\
 \]
By using the property, ${({\log _a}{b^{ - 1}})^m} = {({\log _a}b)^{ - m}}$ we can write,
\[ = {\log _{\dfrac{2}{5}}}{\left( {\dfrac{2}{5}} \right)^{ - 3}}\]
Now we will apply the property ${\log _a}{b^m} = mlo{g_a}b$ to the term
\[ \Rightarrow {\log _{\dfrac{2}{5}}}{\left( {\dfrac{2}{5}} \right)^{ - 3}}\].
\[
\Rightarrow {\log _{\dfrac{2}{5}}}{\left( {\dfrac{2}{5}} \right)^{ - 3}} \\
\Rightarrow ( - 3){\log _{\dfrac{2}{5}}}\left( {\dfrac{2}{5}} \right) \\
 \Rightarrow - 3 \\
 \]
Hence, the value of the expression \[{\log _{\dfrac{2}{5}}}\left( {\dfrac{{125}}{8}} \right)\] is $ - 3$.

Additional Information: A logarithm is the power to which a number must be raised in order to get some other number. Example: ${\log _a}b$ here, a is the base and b is the argument. Exponent is a symbol written above and to the right of a mathematical expression to indicate the operation of raising to a power. The symbol of the exponential symbol is $e$ and has the value $2.17828$. Remember that $\ln a$ and $\log a$ are two different terms. In $\ln a$ the base is e and in $\log a$ the base is $10$. While rewriting an exponential equation in log form or a log equation in exponential form, it is helpful to remember that the base of exponent.

Note: Remember the logarithmic property precisely which is ${({\log _a}{b^{ - 1}})^m} = {({\log _a}b)^{ - m}}$ and ${\log _a}{b^m} = mlo{g_a}b$. While comparing the terms, be cautious. After the application of property when you get the final answer, tress back the problem and see if it returns the same values. Evaluate the base and the argument carefully. Also, remember that ${\ln _e}e = 1$.