Final answer:
To solve the logarithmic equation log₂x + log₂(x+7) = 3, combine the logarithms into a single product and rewrite the equation as 2³ = x(x+7), then solve for x. This results in x = 1, as negative solutions are invalid for logarithms.
Step-by-step explanation:
The question asks us to solve the logarithmic equation log₂x + log₂ (x+7) = 3. To solve this, we use the logarithmic property that the logarithm of a product of two numbers is the sum of the logarithms of the two numbers. So, we can combine the two logarithms on the left hand side: log₂(x(x+7)) = 3. Next, we use the property that if logab = c, then ac = b to rewrite the equation as 23 = x(x+7).
Now we solve for x:
- 23 = x(x+7)
- 8 = x² + 7x
- 0 = x² + 7x - 8
- 0 = (x+8)(x-1)
- x = -8 or x = 1
Because a logarithm cannot have a negative argument, -8 is not a valid solution for the original equation. Therefore, x = 1 is the only solution.