determine the point of intersection
y=logbase 10(x-2)
and y=1-logbase 10(x+1)
solve for first x
10^y=(x-2)
x= 10^y +2
solve for the second x
y-1=log(x+1)
10^(y-1)=x+1 or x= 10^(y-1) +1
the x=x, y=y
10^(y-1)+1=10^y +2
remember a^(b-c)= a^b/a^c so
10^y/10 + 1=10^y +2
10^y (-9/10)=1
or -9*10^y=10
log of each side
-9 +y=1
y=10
so, now figure x..
x= 10^(y-1) +1=10^9+1
weird. Check my work.
you are THE BESTT!!
its wrong
you wrote
10^(y-1)+1=10^y +2
its supposed to be
10^(y-1)-1=10^y +2
THE ANSWER IS APPARENTLY 2, log base 10 2
To determine the point of intersection between the two equations y=logbase 10(x-2) and y=1-logbase 10(x+1), you need to set the equations equal to each other and solve for x.
First, set the two equations equal to each other:
logbase 10(x-2) = 1 - logbase 10(x+1)
Next, since logarithms are exponents, rewrite the equation as:
logbase 10(x-2) + logbase 10(x+1) = 1
Using the property of logarithms, you can combine the two logarithms into a single logarithm:
logbase 10((x-2)(x+1)) = 1
Now, rewrite the equation using exponential form:
(x-2)(x+1) = 10^1
Simplify the equation:
(x-2)(x+1) = 10
Expand the equation:
x^2 - x - 2 = 10
Rearrange the equation to bring all terms to one side:
x^2 - x - 12 = 0
Now, we have a quadratic equation that we can solve. Factoring or using the quadratic formula, we find that the equation factors as:
(x + 3)(x - 4) = 0
Setting each factor equal to zero, we get two solutions:
x + 3 = 0 --> x = -3
x - 4 = 0 --> x = 4
Therefore, the two possible values of x that satisfy the original equation are x = -3 and x = 4.
To find the corresponding y-values, substitute the values of x back into either of the original equations:
For x = -3:
y=logbase 10(-3-2) = logbase 10(-5)
y=1-logbase 10(-3+1) = 1-logbase 10(-2)
For x = 4:
y=logbase 10(4-2) = logbase 10(2)
y=1-logbase 10(4+1) = 1-logbase 10(5)
Note: To evaluate logarithms with negative arguments, you may need to use complex numbers or restrict the solutions depending on the context of the problem.