Final answer:
The real solution to the logarithmic system is x = -3. This was obtained by equating the expressions for y, applying logarithmic properties to combine them, converting to exponential form, and verifying that the solution yields positive arguments in the original logarithms.
Step-by-step explanation:
The student's question involves determining the real solutions to a system of logarithmic equations. We have two equations:
- y = log₂(3x+13) - 10
- y = log₂(x+5) - 9
To find the real solutions, we must equate the two expressions for y and solve for x. This involves the property of logarithms that states logₙ a - logₙ b = logₙ(a/b) for any base n. We can apply this property to combine the logarithmic expressions and isolate x.
Once we have a single equation with the logarithm of a quotient, we can convert to exponential form to solve for x. Finally, we check if the solutions are real by substituting them back into the original equations and ensuring that they are defined (that is, the argument of the logarithms must be positive).
Here is the step-by-step solution:
- Equate the two expressions for y: log₂(3x+13) - 10 = log₂(x+5) - 9.
- Add 9 to both sides: log₂(3x+13) - 1 = log₂(x+5).
- Apply the property of logarithms: log₂((3x+13)/(x+5)) = 1.
- Convert to exponential form: (3x+13)/(x+5) = 2.
- Multiply both sides by (x+5) to solve for x: 3x + 13 = 2(x + 5).
- Expand and simplify: 3x + 13 = 2x + 10.
- Subtract 2x from both sides to find the value of x: x = -3.
- Check if x = -3 is a valid solution by substituting it back into the expressions: log₂(3(-3)+13) and log₂((-3)+5).
Since the arguments of both logarithms with x = -3 are positive, x = -3 is a valid real solution to the system.