Exercise 78 Page 498

Let's combine the solutions above to determine the interval where the equation is defined.

The intersection of the above intervals, which is the interval where the equation is defined, is x>0.

Solving the Equality

We want to solve an equation involving more than one logarithm. To do so, we will use two Properties of Logarithms. Quotient:& log_b mn = log_b m - log_b n Power:& log_b m^n = n log_b m First, we will isolate the logarithm that contains the variable. Then, we can use the above properties to isolate the variable from the logarithm.

Next, 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^2-4x-12=0 ⇔ 1x^2+( - 4)x+( - 12)=0 We see that a= 1, b= - 4, 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=2 ± 4. Therefore, the solutions are x_1=6 and x_2=- 2. However, we have to remember that we can take into consideration only x>0, because only for those x-values the equation is defined. Finally, the solution to the given equation is x=6.