determine the point of intersection
y=logbase 10(x-2)
and y=1-logbase 10(x+1)
A point of intersection is when the two equations equal each other. So set:
log(x-2) = 1 - log(x+1)
Then solve for x.
Does that help, or are the logarithms themselves giving you trouble?
log(x-2)= 1-log(x+1)
log(x-2)+log(x+1)=1
log(a)+log(b)=log(a*b)
log(x-2)+log(x+1)=log((x-2)*(x-1))
1=log(10)
log((x-2)*(x-1))=log(10)
So:
(x-2)*(x+1)=10
x^2-2x+x-2=10
x^2-x-2=10
x^2-x-2-10=0
x^2-x-12=0
Exact solutions of this equation is:
-3 and 4
Negative numbers have not logarithms.
So solution is x=4
To determine the point of intersection between the two given equations, we need to set them equal to each other:
log base 10(x-2) = 1 - log base 10(x+1)
To solve this equation, we can use the properties of logarithms.
First, let's simplify the equation using the logarithmic property log(a) - log(b) = log(a/b):
log base 10[(x-2)/(x+1)] = 1
Next, let's rewrite the equation in exponential form by converting the equation from logarithmic form to exponential form. In logarithmic form, log base b(y) = x means that b^x = y. So, in our case:
[(x-2)/(x+1)] = 10^1
Simplifying further:
(x-2)/(x+1) = 10
Next, let's cross-multiply:
x - 2 = 10(x + 1)
Expand and solve for x:
x - 2 = 10x + 10
-9x = 12
x = -12/9
Finally, substitute the value of x back into one of the original equations to find the corresponding y-coordinate. Let's use the first equation:
y = log base 10(x-2)
y = log base 10(-12/9 - 2)
y = log base 10((-12 - 2 * 9)/9)
y = log base 10(-12 - 18)/9)
y = log base 10(-30)/9)
The point of intersection is (-12/9, log base 10(-30)/9).