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Question: How do you solve \(3{e^x} + 1 = 5\)?...

How do you solve 3ex+1=53{e^x} + 1 = 5?

Explanation

Solution

In this question, we want to solve the expression 3ex+1=53{e^x} + 1 = 5. 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 solution:
In this question, we want to solve the expression,
⇒3ex+1=5\Rightarrow 3{e^x} + 1 = 5
Now, let us simplify the above expression.
For that subtract 1 on both sides.
⇒3ex+1−1=5−1\Rightarrow 3{e^x} + 1 - 1 = 5 - 1
The answer of subtraction of 1 and 1 is 0 from the left –hand side and the answer of subtraction of 5 and 1 is 4 from the right–hand side of the above expression.
Therefore,
⇒3ex=4\Rightarrow 3{e^x} = 4
Now, divide by 3 into both sides.
⇒3ex3=43\Rightarrow \dfrac{{3{e^x}}}{3} = \dfrac{4}{3}
Let us simplify the left-hand side of the above expression.
Division of 3 and 3 is equal to 1.
⇒ex=43\Rightarrow {e^x} = \dfrac{4}{3}
Now, let us take the natural logarithm on both sides to remove the variable from the exponent.
Therefore,
⇒ln⁡(ex)=ln⁡(43)\Rightarrow \ln \left( {{e^x}} \right) = \ln \left( {\dfrac{4}{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⁡(43)\Rightarrow x\ln \left( e \right) = \ln \left( {\dfrac{4}{3}} \right)
Now, we already know that the natural logarithm of e is equal to 1.
Therefore,
⇒x(1)=ln⁡(43)\Rightarrow x\left( 1 \right) = \ln \left( {\dfrac{4}{3}} \right)
Multiply x with 1. That is equal to,
⇒x=ln⁡(43)\Rightarrow x = \ln \left( {\dfrac{4}{3}} \right)

Hence, the value of x is ln⁡(43)\ln \left( {\dfrac{4}{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)