How do you solve $2={{e}^{5x}}$?
Answer
610.8k+ views
Hint: First write the given equation in the order ${{e}^{5x}}=2$. Then take natural log on both the sides to remove the ‘e to the power’ part. Then do the necessary simplification to get the value of ‘x’ by putting the value of $\ln 2$ at last.
Complete step-by-step answer:
Solving the equation means, we have to find the value of ‘x’ for which the equation gets satisfied.
Considering our equation $2={{e}^{5x}}$
It can be written as ${{e}^{5x}}=2$
Taking natural log both the sides, we get
$\Rightarrow \ln {{e}^{5x}}=\ln 2$
As, we know from the logarithmic formula $\ln {{e}^{a}}=a$
So, our equation can be further simplified as
$\Rightarrow 5x=\ln 2$
Dividing both the sides by ‘5’, we get
$\Rightarrow \dfrac{5x}{5}=\dfrac{\ln 2}{5}$
Cancelling out ‘5’ both from the numerator and the denominator, we get
$\Rightarrow x=\dfrac{\ln 2}{5}$
Again as we know, the value of $\ln 2=0.693$
So putting the value of $\ln 2$ in the above equation, we get
$\begin{align}
& \Rightarrow x=\dfrac{0.693}{5} \\
& \Rightarrow x=0.13863 \\
\end{align}$
This is the required solution of the given question.
Note: Taking the natural log on both the sides should be the first approach for solving such questions. Some basic logarithmic rules and values should be known for maximum simplification. For example, $x=\dfrac{\ln 2}{5}$ could be the solution of the given equation, but the value of ‘x’ that we got by putting the value of $\ln 2$ i.e. $x=0.13863$ is more appropriate.
Some basic logarithmic values should be remembered for faster and accurate calculations:
$\begin{align}
& \ln 1=0 \\
& \ln 2=0.693 \\
& \ln 3=1.098 \\
& \ln 4=1.386 \\
& \ln 5=1.609 \\
\end{align}$
Complete step-by-step answer:
Solving the equation means, we have to find the value of ‘x’ for which the equation gets satisfied.
Considering our equation $2={{e}^{5x}}$
It can be written as ${{e}^{5x}}=2$
Taking natural log both the sides, we get
$\Rightarrow \ln {{e}^{5x}}=\ln 2$
As, we know from the logarithmic formula $\ln {{e}^{a}}=a$
So, our equation can be further simplified as
$\Rightarrow 5x=\ln 2$
Dividing both the sides by ‘5’, we get
$\Rightarrow \dfrac{5x}{5}=\dfrac{\ln 2}{5}$
Cancelling out ‘5’ both from the numerator and the denominator, we get
$\Rightarrow x=\dfrac{\ln 2}{5}$
Again as we know, the value of $\ln 2=0.693$
So putting the value of $\ln 2$ in the above equation, we get
$\begin{align}
& \Rightarrow x=\dfrac{0.693}{5} \\
& \Rightarrow x=0.13863 \\
\end{align}$
This is the required solution of the given question.
Note: Taking the natural log on both the sides should be the first approach for solving such questions. Some basic logarithmic rules and values should be known for maximum simplification. For example, $x=\dfrac{\ln 2}{5}$ could be the solution of the given equation, but the value of ‘x’ that we got by putting the value of $\ln 2$ i.e. $x=0.13863$ is more appropriate.
Some basic logarithmic values should be remembered for faster and accurate calculations:
$\begin{align}
& \ln 1=0 \\
& \ln 2=0.693 \\
& \ln 3=1.098 \\
& \ln 4=1.386 \\
& \ln 5=1.609 \\
\end{align}$
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