Question

Express as a single logarithm:
log(x^2+6x+8)-log(x^2-16)

Answer 1

Do you know the log rule:\[\ln (a)-\ln(b) = \ln(\frac{ a }{ b })\]

Answer 2

im learning it, I am a little uncertain as to how to set it up.

Answer 3

Here's a simple example of how to apply the rule above:

\(\log (x^2)-\log(x) = \log(\dfrac{ x^2 }{ x }) = \log x\)

Now apply the rule to your problem.
Then factor the numerator and denominator and reduce the fraction.

Answer 4

x+2/x?

Answer 5

What happened to "log"?

Answer 6

Not quite, its (x+2) for the numerator, on the top, but its not x for the denominator. how did you start?

Answer 7

log=(x+2)/(x-4)?

Answer 8

\(\log(x^2+6x+8)-\log(x^2-16)\)

\(= \log \dfrac{x^2+6x+8}{x^2-16}\)

Now factor the numerator and denominator.
What do you get?

Answer 9

num i got (x+4)(x+2)

Answer 10

\(= \log \dfrac{x^2+6x+8}{x^2-16}\)

\(= \log \dfrac{(x + 4)(x + 2)}{(x+4)(x-4)}\)

\(= \log \dfrac{\cancel{(x + 4)}(x + 2)}{\cancel{(x+4)}(x-4)}\)

\(= \log \dfrac{x + 2}{x - 4}\)

Answer 11

Thanks man you ROCK! I understand clearly now

Answer 12

You're welcome.