Question

How do you solve $7 - 2{e^x} = 5$ ?

Answer 1

Hint : In order to determine the solution of the given equation, simply it first by combining the like terms. 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 exponential value with ${b^y} = x$ form . And write it into logarithmic form. Remember natural logarithm of 1 is zero.

Complete step-by-step answer :
We are given an exponential equation in variable $x$
$7 - 2{e^x} = 5$
Now combining like terms on both of the sides. Terms having $t$ will on the right-Hand side of the equation and constant terms on the left-hand side.
$7 - 5 = 2{e^x} \\\ 2 = 2{e^x} \;$
Dividing both side of the equation with number 2, we get
$\dfrac{2}{2} = \dfrac{{2{e^x}}}{2} \\\ 1 = {e^x} \\\ {e^x} = 1 \;$
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 ${e^x} = 1$ ,comparing it with ${b^y} = X$ , we get values of variables as
$X = 1 \\\ b = e \\\ y = x \;$
So we have logarithmic form of our equation as
$\Rightarrow {\log _e}\left( 1 \right) = x$
Since logarithm with base as $e$ is simply written as $\ln$ and known as” natural logarithm”.
$\Rightarrow \ln \left( 1 \right) = x$
Since natural logarithm of constant 1 is equal to zero, so
$\Rightarrow x = 0$
Therefore, the solution of the given equation is $x = 0$ .
So, the correct answer is “x=0”.

Note : 1. Don’t forgot to cross check your result
2. $\ln$ is known as natural logarithm – Logarithm having base as $e$
3.Logarithm of constant 1 is equal to zero.
4.Exponent and logarithm are basically inverse of each other.
5. The value of exponential constant $e$ is $2.71828$ .
6. Like terms are the terms having the same variable and power. Coefficient of the terms might be different.