Question

How do you solve $7 = 5{e^{0.2x}}$ ?

Answer 1

Hint : In order to determine solution of the given equation, simply it first by dividing both sides by $5$ .Now We will convert the expression into logarithmic form, and to do so use the definition of exponents that the exponent of the form ${b^y} = x$ is when converted into logarithmic form is equivalent to ${\log _b}X = y$ ,so compare with the given logarithm value with this form . And write it into logarithmic form and solve the equation for $x$ to get the required result.

Complete step-by-step answer :
We are given an exponential equation in variable $x$
$7 = 5{e^{0.2x}}$
Simplifying the equation by dividing both sides of the equation with $5$ , we get
$\dfrac{1}{5} \times 7 = \dfrac{1}{5} \times 5{e^{0.2x}} \\\ \dfrac{7}{5} = {e^{0.2x}} \;$
Now we are going to convert the above exponential form into logarithmic form.
Any exponential form ${b^y} = X$ when converted into equivalent logarithmic form results in ${\log _b}X = y$
So in our case we have $\dfrac{7}{5} = {e^{0.2x}}$ ,comparing it with ${b^y} = X$ , we get values of variables as
$X = \dfrac{7}{5} \\\ b = e \\\ y = 0.2x \;$
So we have logarithmic form of our equation as
$\Rightarrow {\log _e}\dfrac{7}{5} = 0.2x$
Since logarithm with base as $e$ is simply written as $\ln$ and known as” natural logarithm”.
$\Rightarrow \ln \dfrac{7}{5} = 0.2x$
Now dividing both sides of the equation with $0.2$ , we have value of $x$ as
$\Rightarrow \dfrac{1}{{0.2}}\ln \dfrac{7}{5} = \dfrac{{0.2x}}{{0.2}} \\\ \Rightarrow 5\ln \dfrac{7}{5} = x \\\ \Rightarrow x = 5\ln \dfrac{7}{5} \;$
Therefore, the solution of the given equation is $x = 5\ln \dfrac{7}{5}$ .
So, the correct answer is “$x = 5\ln \dfrac{7}{5}$”.

Note : 1. A logarithm is basically the reverse of a power or we can say when we calculate a logarithm of any number , we actually undo an exponentiation.
2.Any multiplication inside the logarithm can be transformed into addition of two separate logarithm values .
${\log _b}(mn) = {\log _b}(m) + {\log _b}(n)$
3. Any division inside the logarithm can be transformed into subtraction of two separate logarithm values .
${\log _b}(\dfrac{m}{n}) = {\log _b}(m) - {\log _b}(n)$
4. Any exponent value on anything inside the logarithm can be transformed and moved out of the logarithm as a multiplier and vice versa.
$n\log m = \log {m^n}$