Question

How do you solve $3{e^x} - 2 = 0$?

Answer 1

Hint: In this question, we want to solve the expression $3{e^x} - 2 = 0$. Here, simplify the expression in the form of ${e^x}$. Now, to solve the given expression we will use a natural logarithm to remove variable x from the exponent. Then apply the formula $\ln \left( {{e^x}} \right) = x\ln \left( e \right)$. And the natural logarithm of e is equal to 1.

Complete step-by-step answer:
In this question, we want to solve the expression,
$ \Rightarrow 3{e^x} - 2 = 0$
Now, let us simplify the above expression.
For that add 2 on both sides.
$ \Rightarrow 3{e^x} - 2 + 2 = 0 + 2$
Let us solve the left –hand side of the above expression.
Subtraction of 2 and 2 is 0.
And on the right-hand side, the addition of 0 and 2 is 2.
Therefore,
$ \Rightarrow 3{e^x} = 2$
Now, divide by 3 on both sides.
$ \Rightarrow \dfrac{{3{e^x}}}{3} = \dfrac{2}{3}$
Let us simplify the left-hand side of the above expression.
Division of 3 and 3 is equal to 1.
$ \Rightarrow {e^x} = \dfrac{2}{3}$
Now, let us take the natural logarithm on both sides to remove the variable from the exponent.
Therefore,
$ \Rightarrow \ln \left( {{e^x}} \right) = \ln \left( {\dfrac{2}{3}} \right)$
Now, expand the left-hand side by applying the formula: $\ln \left( {{e^x}} \right) = x\ln \left( e \right)$
Apply this formula on the left-hand side of the expression.
$ \Rightarrow x\ln \left( e \right) = \ln \left( {\dfrac{2}{3}} \right)$
Now, we already know that the natural logarithm of e is equal to 1.
Therefore,
$ \Rightarrow x\left( 1 \right) = \ln \left( {\dfrac{2}{3}} \right)$
Multiply x with 1. That is equal to,
$ \Rightarrow x = \ln \left( {\dfrac{2}{3}} \right)$

Hence, the value of x is $\ln \left( {\dfrac{2}{3}} \right)$ for the given expression$3{e^x} - 2 = 0$.

Note:
 The natural logarithm of a number is its logarithm to the base of the mathematical constant e, where e is an irrational number. And the value of e is approximately equal to 2.718281828459.
Some properties we have to remember are as below.
$\ln 1 = 0$
$\ln e = 1$
$\ln \left( {xy} \right) = \ln x + \ln y$
$\ln \left( {{x^y}} \right) = y\ln \left( x \right)$