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

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} $