Question

How do you solve ${{e}^{-x}}=60$ ?

Answer 1

Hint: To solve the above equation ${{e}^{-x}}=60$ means we have to find the value of x. Now, to find the value of x, we are going to take ${{\log }_{e}}$ or $\ln $ both the sides of the equation. Then we are going to use the property of the logarithm which states that $\ln {{e}^{a}}=a$.

Complete step-by-step answer:
The equation given in the above question which we have to solve is as follows:
${{e}^{-x}}=60$
Solution of the above question is “x” so we have to find the value of x. Now, to get the value of x we have to eliminate this $e$ which can be done by taking ${{\log }_{e}}$ or $\ln $ on both the sides of the above equation and we get,
$\Rightarrow \ln {{e}^{-x}}=\ln 60$ ……….. (1)
We know the logarithm property of the exponent which states that:
$\ln {{e}^{a}}=a$
Using the above relation in the L.H.S of eq. (1) we get,
$\begin{align}
  & \ln {{e}^{-x}}=\ln 60 \\
 & \Rightarrow -x=\ln 60 \\
\end{align}$
Now, we are going to find the value of $\ln 60$. For that we are going to write $\ln 60$ as 2.303 multiplied by ${{\log }_{10}}60$ and we get,
$-x=2.303{{\log }_{10}}60$
We know there is a logarithmic property which states that:
$\log ab=\log a+\log b$
Applying the above property we get,
$\begin{align}
  & -x=2.303{{\log }_{10}}\left( 6\times 10 \right) \\
 & \Rightarrow -x=2.303\left( {{\log }_{10}}6+{{\log }_{10}}10 \right) \\
 & \Rightarrow -x=2.303\left( {{\log }_{10}}6+1 \right) \\
\end{align}$
We know the value of ${{\log }_{10}}6=0.778$ so substituting this value in the above equation we get,
$\begin{align}
  & \Rightarrow -x=2.303\left( 0.778+1 \right) \\
 & \Rightarrow -x=2.303\left( 1.778 \right) \\
 & \Rightarrow x=-4.095 \\
\end{align}$
Hence, we have calculated the value of x in the above equation as -4.095.

Note: From the above problem, we have learnt an interesting thing in which we have learnt how to convert $\ln x$ to $\log x$ which is equal to:
$\ln x=2.303\log x$
So, if in any problem we have given $\ln x$ then we can convert into ${{\log }_{10}}$ form. Because generally, we remember the value of ${{\log }_{10}}$ so it will be easy to find the value of $\ln x$.