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If two equivalent logarithmic expressions have the same base, then the arguments must be equal.
x ≈ 6.04
We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms. log_b mn = log_b m + log_b n First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm. Note that the property holds also for the natural logarithm.
ln(a) + ln(b)=ln(ab)
Distribute x
ln(e^a) = a
Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal.
We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. x^2+3x-e^4=0 ⇕ 1x^2+ 3x+( - e^4)=0 We see that a= 1, b= 3, and c= - e^4. Let's substitute these values into the Quadratic Formula.
Substitute values
Calculate power
a * 1=a
- a(- b)=a* b
Using the Quadratic Formula, we found that the solutions of the given equation are x= -3 ± sqrt(9+4e^4)2. Therefore, the exact solutions are x_1= -3 +sqrt(9+4e^4)2 and x_2= -3 - sqrt(9+4e^4)2. Let's use a calculator to approximate these values. x_1 & = -3 + sqrt(9+4e^4)/2 ≈ 6.04 x_2 & =-3 - sqrt(9+4e^4)/2 ≈ - 9.04 To check for extraneous solutions, we will substitute both 6.04 and - 9.04 for x in the given equation one at a time.
x= 6.04
Add terms
Calculate logarithm
Add terms
Since substituting 6.04 for x in the given equation produces a true statement is a solution to our equation. Let's now check x=-9.04.
x= -9.04
Add terms
The argument of a logarithmic function cannot be negative, so both ln (-9.04) and ln(-6.04) are undefined, and -9.04 is an extraneous solution.