Solve the following logarithmic equation.
$\log _4(s+21)-\log _4(s+5)=\log _4 s$
Select the correct choice below and, if necessary, fill in the answer box.
A. The solution(s) is/are $\square$
(Simplify your answer. Type an integer or a fraction. Use a comma to separate answers as needed.)
B. The solution is not a real number.
Answer (1)
Use the logarithm property to combine the left side of the equation: lo g 4 ( s + 21 ) − lo g 4 ( s + 5 ) = lo g 4 s + 5 s + 21 .
Equate the arguments: s + 5 s + 21 = s .
Rearrange to form a quadratic equation: s 2 + 4 s − 21 = 0 .
Solve the quadratic equation: s = − 7 or s = 3 . Since 0"> s > 0 , the only solution is 3 .
Explanation
Understanding the Problem and Constraints We are given the logarithmic equation lo g 4 ( s + 21 ) − lo g 4 ( s + 5 ) = lo g 4 s . Our goal is to solve for s . Remember that the argument of a logarithm must be positive, so we must have 0"> s + 21 > 0 , 0"> s + 5 > 0 , and 0"> s > 0 . This means that -21"> s > − 21 , -5"> s > − 5 , and 0"> s > 0 . Combining these inequalities, we must have 0"> s > 0 .
Using Logarithm Properties Using the logarithm property lo g b x − lo g b y = lo g b y x , we can rewrite the left side of the equation as lo g 4 ( s + 21 ) − lo g 4 ( s + 5 ) = lo g 4 s + 5 s + 21 . Thus, the equation becomes lo g 4 s + 5 s + 21 = lo g 4 s .
Equating Arguments and Expanding Since the bases of the logarithms are the same, we can equate the arguments: s + 5 s + 21 = s . Multiplying both sides by s + 5 , we get s + 21 = s ( s + 5 ) . Expanding the right side, we have s + 21 = s 2 + 5 s .
Forming and Factoring Quadratic Equation Rearranging the equation to form a quadratic equation, we get s 2 + 5 s − s − 21 = 0 , which simplifies to s 2 + 4 s − 21 = 0. Factoring the quadratic equation, we have ( s + 7 ) ( s − 3 ) = 0.
Solving for s and Checking the Solution Solving for s , we find two possible solutions: s = − 7 or s = 3 . However, we know that s must be greater than 0, so s = − 7 is not a valid solution. Therefore, the only possible solution is s = 3 . Let's check if s = 3 is indeed a solution by substituting it into the original equation: lo g 4 ( 3 + 21 ) − lo g 4 ( 3 + 5 ) = lo g 4 ( 24 ) − lo g 4 ( 8 ) = lo g 4 8 24 = lo g 4 3. Since lo g 4 ( s ) = lo g 4 ( 3 ) , the solution s = 3 is valid.
Final Answer Therefore, the solution to the logarithmic equation is s = 3 .
Examples
Logarithmic equations are used in various fields such as physics, engineering, and finance. For example, in finance, they are used to calculate the time it takes for an investment to double at a certain interest rate. In physics, they are used to describe the decay of radioactive materials. Understanding how to solve logarithmic equations is essential for solving real-world problems in these fields.
