To solve this logarithmic equation, we can use the properties of logarithms to combine the two logarithms on the left side into a single logarithm. The property we will use is the product rule of logarithms, which states that logₐ(x) + logₐ(y) = logₐ(xy).
Given: log₂(2x-18) + log₂(x-9) = 5
Using the product rule of logarithms: log₂((2x-18)(x-9)) = 5
Now we can rewrite the equation in exponential form to solve for x: 2^5 = (2x-18)(x-9)
Now we need to solve this quadratic equation. We can factor it or use the quadratic formula. Let's use the quadratic formula: x = (-(-27) ± √((-27)^2 - 42130)) / (2*2) x = (27 ± √(729 - 1040)) / 4 x = (27 ± √(-311)) / 4
Since the square root of -311 is not a real number, we have no real solutions to this equation.
Therefore, the equation log₂(2x-18) + log₂(x-9) = 5 has no real solutions.
To solve this logarithmic equation, we can use the properties of logarithms to combine the two logarithms on the left side into a single logarithm. The property we will use is the product rule of logarithms, which states that logₐ(x) + logₐ(y) = logₐ(xy).
Given: log₂(2x-18) + log₂(x-9) = 5
Using the product rule of logarithms:
log₂((2x-18)(x-9)) = 5
Now we can rewrite the equation in exponential form to solve for x:
2^5 = (2x-18)(x-9)
32 = 2x^2 - 18x - 9x + 162
32 = 2x^2 - 27x + 162
0 = 2x^2 - 27x + 130
Now we need to solve this quadratic equation. We can factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-(-27) ± √((-27)^2 - 42130)) / (2*2)
x = (27 ± √(729 - 1040)) / 4
x = (27 ± √(-311)) / 4
Since the square root of -311 is not a real number, we have no real solutions to this equation.
Therefore, the equation log₂(2x-18) + log₂(x-9) = 5 has no real solutions.