Solveeit Logo

Question

Question: How do you solve \(2{{e}^{5x+2}}=8\) ?...

How do you solve 2e5x+2=82{{e}^{5x+2}}=8 ?

Explanation

Solution

We are given an expression in which an exponential term is being equated to a constant term, 8. In order to find the solution of the given equation, we must find the value of x. Thus, we shall first make the coefficient of the exponential term equal to 1 by dividing the entire equation by 2. Further we will take logarithms on the left hand side and the right hand side of the equation and simplify the equation more.

Complete step by step solution:
Given that 2e5x+2=82{{e}^{5x+2}}=8.
In order to simplify the present equation, we shall divide it by 2 on the left hand side as well as the right hand side and make the coefficient of the exponential term equal to 1.
e5x+2=82\Rightarrow {{e}^{5x+2}}=\dfrac{8}{2}
e5x+2=4\Rightarrow {{e}^{5x+2}}=4
Now we shall take the logarithm with base e on both sides of the equation.
loge5x+2=log4\Rightarrow \log {{e}^{5x+2}}=\log 4
Here, we will use the property of logarithm, logab=bloga\log {{a}^{b}}=b\log a.
(5x+2)loge=log4\Rightarrow \left( 5x+2 \right)\log e=\log 4
We know that logee=1{{\log }_{e}}e=1, thus substituting this value we get,
(5x+2)1=log4\Rightarrow \left( 5x+2 \right)1=\log 4
Thus, we have obtained a linear equation in x-variable term and 2 constant terms now. We will transpose the constant term 2 to the right hand side and divide both sides of the equation by 5 to find our final solution.
5x=log42\Rightarrow 5x=\log 4-2
x=log425\Rightarrow x=\dfrac{\log 4-2}{5}
Since log4=1.386\log 4=1.386, hence substituting this value, we get
x=1.38625\Rightarrow x=\dfrac{1.386-2}{5}
x=0.6145\Rightarrow x=\dfrac{-0.614}{5}
x=0.1228\Rightarrow x=-0.1228
Therefore, the solution for the given exponential equation 2e5x+2=82{{e}^{5x+2}}=8 is x=log425x=\dfrac{\log 4-2}{5} or x=0.1228x=-0.1228

Note: Another way by which we could have simplified the term log4 and found out its value is by using the logarithmic property logab=bloga\log {{a}^{b}}=b\log a as log4 can also be written as log22\log {{2}^{2}}. Thus, log4=2log2\log 4=2\log 2 and we could have substituted the value of log 2 to find the value of log4.