Precalc help - combining logarithms

[Belinda]

It's about 11 pm right now and I've been working on this assignment for 5 hours. I could really use a good night's sleep and a little help to go with it.

I'm on one of the last equations which asks me to simplify the following by combining all terms into a single logarithm:

ln(x^2)+(1/3)ln(x-2)-4

I'm trying to make sense of the notes from yesterday's lecture but I don't understand which properties I'm supposed to apply first. All my notes are in "log" and this problem is "ln" which is also really confusing me.

My professor isn't very good at clarity or explanation. It's been a long day and I would really appreciate some clear, step by step guidance for solving this.

 

[pappus]

It's about 11 pm right now and I've been working on this assignment for 5 hours. I could really use a good night's sleep and a little help to go with it.

I'm on one of the last equations which asks me to simplify the following by combining all terms into a single logarithm:

ln(x^2)+(1/3)ln(x-2)-4

I'm trying to make sense of the notes from yesterday's lecture but I don't understand which properties I'm supposed to apply first. All my notes are in "log" and this problem is "ln" which is also really confusing me.

My professor isn't very good at clarity or explanation. It's been a long day and I would really appreciate some clear, step by step guidance for solving this.

1. The necessary logarithmic laws to answer this question:
\(\displaystyle \displaystyle{\ln(a) + \ln(b) = \ln(a \cdot b)}\) ........... \(\displaystyle \displaystyle{\ln(a) - \ln(b) = \ln\left(\frac ab \right)}\)

\(\displaystyle \displaystyle{\ln(b^a) = a \cdot \ln(b)}\) From this law follows: \(\displaystyle \displaystyle{\ln(\sqrt[n]{a}) = \frac1n \cdot \ln(a)}\)

2. You are supposed to know that ln(e) = 1. Therefore \(\displaystyle \displaystyle{4=4 \cdot \ln(e) = \ln(e^4)}\)

3. Using these laws you'll get a quotient as the argument of the logarithm whose numerator is a product and whose denominator is a power of e.

 

[Belinda]

1. The necessary logarithmic laws to answer this question:
\(\displaystyle \displaystyle{\ln(a) + \ln(b) = \ln(a \cdot b)}\) ........... \(\displaystyle \displaystyle{\ln(a) - \ln(b) = \ln\left(\frac ab \right)}\)

\(\displaystyle \displaystyle{\ln(b^a) = a \cdot \ln(b)}\) From this law follows: \(\displaystyle \displaystyle{\ln(\sqrt[n]{a}) = \frac1n \cdot \ln(a)}\)

2. You are supposed to know that ln(e) = 1. Therefore \(\displaystyle \displaystyle{4=4 \cdot \ln(e) = \ln(e^4)}\)

3. Using these laws you'll get a quotient as the argument of the logarithm whose numerator is a product and whose denominator is a power of e.

Thanks for the help. Actually I got something completely different. It took me about 15 minutes to figure out how to do it. This was what I got:

1) ln(x^2)+(1/3)(x-2)-4
2) = ln(x^2)+ln((x-2)-4)^1/3)
3) = ln(x^2((x-2)-4)^1/3)

Is this even correct? I wasn't sure what to do with the 4...let me try your way.

 

[Belinda]

If line 1 is correct, line 3 is not.

Which is the problem, to simplify

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4, or\ ln(x^2) + \frac{1}{3}ln[(x - 2) - 4]?\)

By the way, ln is just the standard abbreviation for a log to the base of e.

Oh oops I forgot to put the "ln" there. The original problem is ln(x^2)+(1/3)ln(x-2)-4 so it's the first one that you stated.

 

[Belinda]

OK then. You sorta kinda have the idea

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln(x^2) + ln\left[(x - 2)^{1/3}\right] - 4\) You can't pull the 4 inside the ln(x - 2)^(1/3) because it is NOT in log form

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln(x^2) + ln\left[(x - 2)^{1/3}\right] - 4 = ln\left[x^2(x - 2)^{1/3}\right] - 4\) You got the part about the addition of logs right

Note \(\displaystyle (x - 2)^{1/3} = \sqrt[3]{x - 2}\)

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln\left(x^2\sqrt[3]{x - 2}\right) - 4\)

Now all you have left to deal with is the 4, which pappus previously explained.

Thank you very much. This was my answer, is it right?

 

[Belinda]

OK then. You sorta kinda have the idea

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln(x^2) + ln\left[(x - 2)^{1/3}\right] - 4\) You can't pull the 4 inside the ln(x - 2)^(1/3) because it is NOT in log form

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln(x^2) + ln\left[(x - 2)^{1/3}\right] - 4 = ln\left[x^2(x - 2)^{1/3}\right] - 4\) You got the part about the addition of logs right

Note \(\displaystyle (x - 2)^{1/3} = \sqrt[3]{x - 2}\)

\(\displaystyle ln(x^2) + \frac{1}{3}ln(x - 2) - 4 = ln\left(x^2\sqrt[3]{x - 2}\right) - 4\)

Now all you have left to deal with is the 4, which pappus previously explained.

Thank you. This was what I got: