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Question: How do you solve \(3{e^x} - 2 = 0\)?...

How do you solve 3ex−2=03{e^x} - 2 = 0?

Explanation

Solution

In this question, we want to solve the expression 3ex−2=03{e^x} - 2 = 0. Here, simplify the expression in the form of ex{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⁡(ex)=xln⁡(e)\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,
⇒3ex−2=0\Rightarrow 3{e^x} - 2 = 0
Now, let us simplify the above expression.
For that add 2 on both sides.
⇒3ex−2+2=0+2\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,
⇒3ex=2\Rightarrow 3{e^x} = 2
Now, divide by 3 on both sides.
⇒3ex3=23\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.
⇒ex=23\Rightarrow {e^x} = \dfrac{2}{3}
Now, let us take the natural logarithm on both sides to remove the variable from the exponent.
Therefore,
⇒ln⁡(ex)=ln⁡(23)\Rightarrow \ln \left( {{e^x}} \right) = \ln \left( {\dfrac{2}{3}} \right)
Now, expand the left-hand side by applying the formula: ln⁡(ex)=xln⁡(e)\ln \left( {{e^x}} \right) = x\ln \left( e \right)
Apply this formula on the left-hand side of the expression.
⇒xln⁡(e)=ln⁡(23)\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,
⇒x(1)=ln⁡(23)\Rightarrow x\left( 1 \right) = \ln \left( {\dfrac{2}{3}} \right)
Multiply x with 1. That is equal to,
⇒x=ln⁡(23)\Rightarrow x = \ln \left( {\dfrac{2}{3}} \right)

Hence, the value of x is ln⁡(23)\ln \left( {\dfrac{2}{3}} \right) for the given expression3ex−2=03{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 1 = 0
ln⁡e=1\ln e = 1
ln⁡(xy)=ln⁡x+ln⁡y\ln \left( {xy} \right) = \ln x + \ln y
ln⁡(xy)=yln⁡(x)\ln \left( {{x^y}} \right) = y\ln \left( x \right)