Question

How do you solve $ 5{e^{2x + 11}} = 30 $ ?

Answer 1

Hint: When we are given an equation with one variable and we are asked to solve the equation, it simply means that we have to find the value of the variable (here $ x $ ). In order to find the value of $ x $ , we need to shift everything else except $ x $ itself to RHS, leaving only $ x $ in the LHS and we do this by simple addition, multiplication and division by a constant on the whole equation. Another important formula used in the question is a property of logarithms i.e., if $ {\log _b}a = n $ then $ {b^n} = a $ . Make use of this property to solve the given equation.

Complete step by step solution:
(i)
We are given an equation
 $ 5{e^{2x + 11}} = 30 $
We have to solve this equation for the given variable i.e., $ x $ . In simpler words, we have to find the value of $ x $ for which this equation holds true.
In order to solve this equation, we will only keep the variable in the LHS and everything else in the RHS. So, dividing the equation by $ 5 $ , we will get:
 $
  \dfrac{{5{e^{2x + 11}}}}{5} = \dfrac{{30}}{5} \\
  {e^{2x + 11}} = 6 \;
  $
(ii)
Now, as we know a property of logarithm i.e., if $ {\log _b}a = n $ then, it means $ {b^n} = a $ .
Therefore, we can write $ {e^{2x + 11}} = 6 $ as:
 $ {\log _e}6 = 2x + 11 $
As we know that $ {\log _e}x $ can also be written as $ \ln x $ . Our equation will become:
 $ \ln 6 = 2x + 11 $
(iii)
Now, we have our equation:
 $ 2x + 11 = \ln 6 $
As we know that, in order to find the value of $ x $ we need to keep it in LHS and shift everything else in RHS. We will, subtract $ 11 $ from both the sides:
 $
  2x + 11 - 11 = \ln 6 - 11 \\
  2x = \ln 6 - 11 \;
  $
(iv)
Now, we will divide both the sides by $ 2 $ in order to get $ x $ in the LHS.
 $
  \dfrac{{2x}}{2} = \dfrac{{\ln 6 - 11}}{2} \\
  x = \dfrac{{\ln 6 - 11}}{2} \;
  $
(v)
Now, we will solve the obtained expression. For solving we first need to find the value of $ \ln 6 $ which we will find from the calculator.
So, $ \ln 6 = 1.791 $
Putting this value in our expression, we will have:
 $
  x = \dfrac{{1.791 - 11}}{2} \\
  x = \dfrac{{ - 9.209}}{2} \\
  x = - 4.60 \;
  $
Hence, for $ 5{e^{2x + 11}} = 30 $ the solution is $ x = - 4.60 $
So, the correct answer is “ $ x = - 4.60 $ ”.

Note: We could also solve this question by taking log on both sides of the equation $ {e^{2x + 11}} = 6 $ . In this case, it would have become $ \ln {e^{2x + 11}} = \ln 6 $ and as we know that $ \ln {e^n} $ is $ n $ itself so we would have got $ 2x + 11 = \ln 6 $ . Thus, this approach gives the same result as our approach of directly applying the logarithm property. In case you forget the identity, you can apply this method to solve such questions.