Determine the intersection point of the curves y=log(x-2) and y=1-log(x+1).
Both of them are base 10
I haven't done log for a while, but can't you just use subsitution and set up an equation like this?
log(x-2)=1-log(x+1) and then solve for x?
log(x-2)+log(x+1)=1 and I think there's a rule that let you combine those... like multiplying them or something and go from there. Or if you have a graphing calculator, just use that to find the intersections for checking your answer.
Jake is on the right track, from his
log(x-2)+log(x+1)=1
log((x-2)(x+1))=1
log(x^2 - x - 2) = 1
x^2 - x - 2 = 10^1
x^2 - x - 12 = 0
(x-4)(x+3) = 0
x = 4 or x = -3, but x > -1 for log (x+1) to be defined,so
x = 4
To determine the intersection point of the curves y=log(x-2) and y=1-log(x+1), we first need to set the two equations equal to each other and solve for x.
log(x-2) = 1 - log(x+1)
To combine the logarithms, we can use the properties of logarithms:
log(x-2) + log(x+1) = 1
Next, we can simplify the equation using the rules of logarithms:
log[(x-2)(x+1)] = 1
Now, we can remove the logarithm by converting the equation into exponential form:
10^1 = (x-2)(x+1)
Simplifying further:
10 = x^2 - x - 2x + 2
Rearranging the terms:
x^2 - 3x - 8 = 0
Now, we need to solve this quadratic equation for x. We can factorize it or use the quadratic formula to find the solutions. The factored form of the equation is:
(x + 1)(x - 8) = 0
Setting each factor equal to zero, we have:
x + 1 = 0 or x - 8 = 0
Solving each equation, we find:
x = -1 or x = 8
Therefore, the curves intersect at two points: (-1, log(3)) and (8, log(10-2)).