Mark W. answered • 11/12/14
Tutor
4
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Former Air Force Officer / Certified (SC, TX) HS Math Teacher
Some of the key things to consider here are that the domain of the logarithmic function must be greater than zero, and we cannot divide by zero.
ln(3) is a constant - it just moves the locus ("graph") of the equation, so let's focus on the rest for now.
ln(5+x)-ln(3+x)=ln((5+x)/(3+x))
x may not equal -3, or 3+x would equal zero, and we can't have that. Neither may x equal -5, or the numerator 5+x would equal zero, and we would have to take the logarithm of zero. Finally, x cannot be between -3 and -5, or the quotient (5+x)/(3+x) will be negative, and we would have to take the logarithm of a negative number.
So, the domain is: x<-5 U -3<x (Read that "x is less than negative five union negative 3 is less than x.")
The range is a little more problematic.
Since we are subtracting ln(3) from the result, the locus of the equation will be asymptotic to that value, which is about equal to -1.099. In other words, as the locus of the graph goes off to infinity on the negative x side, the values of y will be negative, but will increase, getting closer and closer to -1.099, but never actually getting there. Similarly, on the right side, as the x's are positive and get bigger, the values of y will approach -1.099 from above, getting smaller and smaller, but never quite reaching this number.
Otherwise, as the graph approaches -5 from the left, the domain will get arbitrarily small (in other words, big negative numbers), and, similarly, as the graph approaches -3 from the right, the domain will get arbitrarily big.
So, the range is y may not equal ln(3).
ln(3) is a constant - it just moves the locus ("graph") of the equation, so let's focus on the rest for now.
ln(5+x)-ln(3+x)=ln((5+x)/(3+x))
x may not equal -3, or 3+x would equal zero, and we can't have that. Neither may x equal -5, or the numerator 5+x would equal zero, and we would have to take the logarithm of zero. Finally, x cannot be between -3 and -5, or the quotient (5+x)/(3+x) will be negative, and we would have to take the logarithm of a negative number.
So, the domain is: x<-5 U -3<x (Read that "x is less than negative five union negative 3 is less than x.")
The range is a little more problematic.
Since we are subtracting ln(3) from the result, the locus of the equation will be asymptotic to that value, which is about equal to -1.099. In other words, as the locus of the graph goes off to infinity on the negative x side, the values of y will be negative, but will increase, getting closer and closer to -1.099, but never actually getting there. Similarly, on the right side, as the x's are positive and get bigger, the values of y will approach -1.099 from above, getting smaller and smaller, but never quite reaching this number.
Otherwise, as the graph approaches -5 from the left, the domain will get arbitrarily small (in other words, big negative numbers), and, similarly, as the graph approaches -3 from the right, the domain will get arbitrarily big.
So, the range is y may not equal ln(3).
