Log(x-2)+log2=2logy
To solve the equation logarithmically, we will use the properties of logarithms. Let's break it down step by step:
1. Start by combining the logarithms:
log(x-2) + log(2) = 2log(y)
2. Apply the logarithmic property of addition:
log((x-2)*2) = log(y^2)
3. Remove the logarithm by equating the expressions inside the log:
(x-2)*2 = y^2
4. Simplify the equation:
2x - 4 = y^2
5. Rearrange the equation to solve for y:
y^2 = 2x - 4
6. Take the square root of both sides:
y = ±√(2x - 4)
So, the solution for y is y = ±√(2x - 4).
To check the validity of this solution, substitute the value of y back into the original equation: log(x-2) + log(2) = 2log(y). If both sides of the equation are equal, then the solution is valid.
I will assume you want this simplified
Using basic log rules:
Log(x-2)+log2=2logy
log( (x-2)(2) ) = log (y^2)
anti-log it
2(x-2) = y^2
y^2 = 2x - 4 or x = (y^2 + 4)/2 or ....