1. Solve: log (y-1) = -log(y-9)
2. Find the inverse of the logarithmic function f defined by f(y)= 2log5 (2y-8)+3.
1. To solve the equation log(y-1) = -log(y-9), we can start by using the properties of logarithms.
First, we can apply the property log(a) = -log(b) to rewrite the equation as log(y-1) + log(y-9) = 0.
Next, we can combine the logarithms using the property log(a) + log(b) = log(ab). So we have log((y-1)(y-9)) = 0.
Since the logarithm is equal to 0, the argument inside the logarithm must be equal to 1. So we have (y-1)(y-9) = 1.
Expanding the equation, we get y^2 - 10y + 9 = 1. Rearranging terms, we have y^2 - 10y + 8 = 0.
This is a quadratic equation, which we can solve using factoring, completing the square, or the quadratic formula. In this case, let's factor it.
The factors of 8 that add up to -10 are -2 and -8. So we have (y-2)(y-8) = 0.
Therefore, the solutions are y = 2 and y = 8.
2. To find the inverse of the logarithmic function f defined by f(y) = 2log5(2y-8) + 3, we need to switch the roles of x and y.
Let's start by writing the equation in terms of x instead of y. The given function is f(x) = 2log5(2x-8) + 3.
To find the inverse, we want to solve for x when f(x) = y. So we'll start by setting y equal to the equation of the function: y = 2log5(2x-8) + 3.
Next, we'll isolate the logarithm term, subtracting 3 from both sides: y - 3 = 2log5(2x-8).
Now, divide both sides by 2 to isolate the log term: (y - 3)/2 = log5(2x-8).
To eliminate the logarithm, we need to rewrite the equation in exponential form. Since log5(2x-8) represents that 5 is raised to what power equals (2x-8), we rewrite it as 5^((y-3)/2) = 2x - 8.
To isolate x, we'll first add 8 to both sides: 5^((y-3)/2) + 8 = 2x.
Finally, divide both sides by 2 to get x by itself: x = (5^((y-3)/2) + 8)/2.
Therefore, the inverse function is given by g(y) = (5^((y-3)/2) + 8)/2.