express a single natural log: ln(x-3)+ln(x+3)
a. ln(x^2-9)
b. -ln6
c. ln2x
d. ln(2x-9)
e. none
Using
log(xy) = log x + log y
the answer is (a):
ln(x-3)+ln(x+3) = ln[(x-3)(x+3)] = ln(x^2-3x+3x-9) = ln(x^2-9)
You will be using the following property to solve this expression:
ln(a) + ln(b)=ln(ab)
Your "a" is x-3 and your "b" is x+3. So, the expression can be combined as follows:
ln(x-3)+ln(x+3)
ln[(x+3)(x-3)]
Now you need to follow the FOIL process:
ln(x^2 - 3x + 3x -9)
ln(x^2 - 9)
The answer is ln(x^2 - 9), or letter A.
using the laws of logs, by adding them, you can multiply what you have in the brackets to get:
ln((x-3)(x+3)) = ln(x^2 - 9)
a
Comments
Using
log(xy) = log x + log y
the answer is (a):
ln(x-3)+ln(x+3) = ln[(x-3)(x+3)] = ln(x^2-3x+3x-9) = ln(x^2-9)
You will be using the following property to solve this expression:
ln(a) + ln(b)=ln(ab)
Your "a" is x-3 and your "b" is x+3. So, the expression can be combined as follows:
ln(x-3)+ln(x+3)
ln[(x+3)(x-3)]
Now you need to follow the FOIL process:
ln(x^2 - 3x + 3x -9)
ln(x^2 - 9)
The answer is ln(x^2 - 9), or letter A.
using the laws of logs, by adding them, you can multiply what you have in the brackets to get:
ln((x-3)(x+3)) = ln(x^2 - 9)
a