Question

How do you solve $100{{e}^{-0.6x}}=20$?

Answer 1

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.