How do you solve \[\left( {\dfrac{1}{4}} \right){e^{ - 2t}} = 0.1\] ?

Hint: First we will apply the rule ${\log _a}a = 1$. Then we will evaluate all the required terms. Then we will apply the property ${\log _a}a = 1$. The value of the logarithmic function $\ln e$ is $1$.

Complete step-by-step solution:
We will first apply the base rule. This rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$, and $x$ is greater than $0$.
\[\left( {\dfrac{1}{4}} \right){e^{ - 2t}} = 0.1\]
Now multiply both the sides by $4$.
\[
  4.\left( {\dfrac{1}{4}} \right){e^{ - 2t}} = 4\left( {0.1} \right) \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,{e^{ - 2t}} = \left( {0.4} \right) \\
 \]
Now we take log on both sides.
\[\ln {e^{ - 2t}} = \ln \left( {0.4} \right)\]
Now apply the logarithmic property ${\log _a}a = 1$. Hence, the equation becomes,
\[ - 2t = \ln \left( {0.4} \right)\]
Now divide both the sides by $ - 2$.
\[
  \dfrac{{ - 2t}}{{ - 2}} = \dfrac{{\ln \left( {0.4} \right)}}{{ - 2}} \\
  \,\,\,\,\,\,\,t = \dfrac{{\ln \left( {0.4} \right)}}{{ - 2}} \\
  \,\,\,\,\,\,t = 0.458145 \\
 \]
Hence, the value of the expression \[\left( {\dfrac{1}{4}} \right){e^{ - 2t}} = 0.1\] is \[0.458145\].

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 ${\log _a}a = 1$.
 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 ${\ln _e}e = 1$.