single logarithm?

Question

express in a single logarithm

ln(a + b) + ln(a - b) - 3 ln(c)
e^(a+b)+e^(a-b)-3e^(c)
(a+b)(a-b) cancel each other?
leaving 3e^(-c)?
or is my thought not correct?

Answer 1

Neither of your steps are correct.
You cannot go from ln(a + b) + ln(a - b) to e^(a+b)+e^(a-b).

The trick is to use the following facts about logarithms:
ln x + ln y = ln (xy)
ln x - ln y = ln (x/y)
ln (r^s) = s ln r

So ln(a + b) + ln(a - b) = ln [(a+b)(a-b)] = ln (a^2 - b^2)
and 3 ln(c) = ln (c^3)

Hence ln(a + b) + ln(a - b) - 3 ln(c)
= ln (a^2 - b^2) - ln (c^3)
= ln [ (a^2 - b^2)/c^3 ]



Edit: Oh, I see the same thing has been posted above. Oh well.

Answer 2

I'm afraid that several of your thoughts are not correct. You were asked to express the results in a single logarithm. That DIDN'T mean exponentiate it; NOR can you separately add or subtract the exponentials of logarithms that are added or subtracted; NOR do (a + b) and (a - b) "cancel each other"; NOR is -3e^(c) equal to 3e^(-c). That is quite aplethora of mathematical mistakes!

Here's the correct result:

ln(a + b) + ln(a - b) - 3 ln(c) = ln [(a^2 - b^2)/c^3].

That's because (i) ln x + ln y = ln (xy). When x = (a + b) and y = (a - b), xy = (a + b)(a - b) = (a^2 - b^2).

Also, (ii) "- 3 ln c" means "divide by the ln (natural logarithm) of c^3."

Hence the result given.

Live long and prosper.

Answer 3

you'd solve a = e^x by taking ln of both sides, getting x = ln a, but your problem is NOT AN EQUATION. It's an expression to simplify using the laws of logarithms:

1. log a + log b = log(ab)
2. log a - log b = log(a/b)
3. a log b = log b^a.

so ln(a+b) + ln(a-b) = ln[(a+b)(a-b)] by rule 1, and that can be written as ln(a²-b²).
3ln(c) = ln(c³) by rule 3, and finally, using rule 2, you get

ln[(a²-b²)/c³], a single log as the question asked and as Dr. Spock answered.