To solve the equation log7(x-1) + log7(x-2) = log7(x+2), we can combine the two logarithms on the left side using the product rule of logarithms, which states that log(a) + log(b) = log(ab).
So, log7((x-1)(x-2)) = log7(x+2)
Now, we can drop the logarithms since they have the same base and the equation becomes:
(x-1)(x-2) = x+2
Expanding the left side:
x^2 - 3x + 2 = x + 2
Rearranging the terms:
x^2 - 4x = 0
Now, we can factor out an x:
x(x - 4) = 0
So, the solutions are x = 0 and x = 4. However, we need to check if these solutions are valid by plugging them back into the original equation:
When x = 0:
log7(0-1) + log7(0-2) = log7(0+2 This simplifies to log7(-1) + log7(-2) = log7(2 Since the logarithm of a negative number is undefined, x = 0 is not a valid solution.
When x = 4:
log7(4-1) + log7(4-2) = log7(4+2 This simplifies to log7(3) + log7(2) = log7(6 This is a valid solution.
Therefore, the only valid solution to the equation log7(x-1) + log7(x-2) = log7(x+2) is x = 4.
To solve the equation log7(x-1) + log7(x-2) = log7(x+2), we can combine the two logarithms on the left side using the product rule of logarithms, which states that log(a) + log(b) = log(ab).
So, log7((x-1)(x-2)) = log7(x+2)
Now, we can drop the logarithms since they have the same base and the equation becomes:
(x-1)(x-2) = x+2
Expanding the left side:
x^2 - 3x + 2 = x + 2
Rearranging the terms:
x^2 - 4x = 0
Now, we can factor out an x:
x(x - 4) = 0
So, the solutions are x = 0 and x = 4. However, we need to check if these solutions are valid by plugging them back into the original equation:
When x = 0:
log7(0-1) + log7(0-2) = log7(0+2
This simplifies to
log7(-1) + log7(-2) = log7(2
Since the logarithm of a negative number is undefined, x = 0 is not a valid solution.
When x = 4:
log7(4-1) + log7(4-2) = log7(4+2
This simplifies to
log7(3) + log7(2) = log7(6
This is a valid solution.
Therefore, the only valid solution to the equation log7(x-1) + log7(x-2) = log7(x+2) is x = 4.