If two equivalent logarithmic expressions have the same base, then the arguments must be equal.
- 4, 3
Practice makes perfect
Before we begin, remember that the argument of a logarithm has to be greater than 0. With that in mind, let's start by excluding any values of x that would result in taking the logarithm of a number less than or equal to zero.
log_8 (x^2+x)=log_8 12
After we determine the range of values on which the equation is defined, we can solve the equation using the Properties of Logarithms and combine our solution sets.
Determining Where the Equation is Defined
For the logarithm to be defined, x^2+x needs to be greater than zero.
Therefore, the equation is defined for x< - 1 or x>0.
Solving the Equation
We want to solve an equation involving more than one logarithm. First, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal.
log_b x=log_b y ⇔ x= y
Let's apply the above property to our equation.
We will use the Quadratic Formula to solve the above quadratic equation.
ax^2+ bx+ c=0 ⇔ x=- b± sqrt(b^2-4 a c)/2 a
we can identify the values of a, b, and c.
x^2+x-12=0 ⇔ 1x^2+( 1)x+( - 12)=0
We see that a= 1, b= 1, and c= - 12. Let's substitute these values into the Quadratic Formula.
Using the Quadratic Formula, we found that the solutions of the given equation are x=- 1 ± 7/2. Therefore, the solutions are x_1=3 and x_2=- 4. Both these values are in the domain of our given logarithm. Finally, the solution to the given equation is x=- 4 and x=3..
Checking Our Answer
Checking the answer
To check our answer, we will substitute - 4 and 3 for x in the given equation. Let's consider one at a time. We will start with substituting - 4 for x.
