Question

How do you solve \[100{{e}^{-0.6x}}=20\]?

Answer 1

Hint: This problem can be solved by logarithmic properties. We know that if a=b then by applying log on both sides we can get \[\log a=\log b\]. By using this property, we can solve the equation to find the values of x.

Complete step by step solution:
For the given problem we are given to solve the equation \[100{{e}^{-0.6x}}=20\].For that let us consider the given equation as equation (1).
\[100{{e}^{-0.6x}}=20................\left( 1 \right)\]
Let us divide the equation (1) with 100 on both sides, we get
\[\Rightarrow \dfrac{100{{e}^{-0.6x}}}{100}=\dfrac{20}{100}\]
By simplifying the above equation, we get
\[\Rightarrow {{e}^{-0.6x}}=\dfrac{1}{5}\]
Let us consider the above equation as equation (2).
\[\Rightarrow {{e}^{-0.6x}}=\dfrac{1}{5}..................\left( 2 \right)\]
Applying \[{{\log }_{e}}\] on both sides of equation (2), we get
\[{{\log }_{e}}\left( {{e}^{-0.6x}} \right)={{\log }_{e}}\left( \dfrac{1}{5} \right)\]
Let us consider the above equation as equation (3).
\[{{\log }_{e}}\left( {{e}^{-0.6x}} \right)={{\log }_{e}}\left( \dfrac{1}{5} \right)..................\left( 3 \right)\]
As we know\[{{\log }_{e}}{{e}^{x}}=x\], therefore let us apply this formula to equation (3).
Let us consider the formula as (f1).
\[{{\log }_{e}}{{e}^{x}}=x...................\left( f1 \right)\]
Let us apply formula (f1) to the equation (3), we get
\[\Rightarrow -0.6x={{\log }_{e}}\left( \dfrac{1}{5} \right)\]
Let us consider the above equation as equation (4).
\[\Rightarrow 0.6x={{\log }_{e}}\left( \dfrac{1}{5} \right)...................\left( 4 \right)\]
As we know that \[{{\log }_{e}}\left( \dfrac{1}{5} \right)=-1.6\] let us apply this to the equation (4), we get
\[\Rightarrow -0.6x=-1.6\]
By simplifying a bit in the equation (4), we get
\[\Rightarrow 0.6x=1.6\]
By dividing the above equation with 0.6, we get
\[\Rightarrow \dfrac{0.6x}{0.6}=\dfrac{1.6}{0.6}\]
By simplifying the equation, we get
\[x=2.67\]
Hence, by solving \[100{{e}^{-0.6x}}=20\] we get \[x=2.67\].

Note: Students should avoid calculation mistakes while solving this problem. If a small mistake is done, then the final answer may get interrupted. Students may have a misconception that \[{{\log }_{e}}{{e}^{x}}=e{}^{x}\] but we know that \[{{\log }_{e}}{{e}^{x}}=x\]. If this misconception is followed, then the final answer may get interrupted.