Question

Write 2 lnx-1/2ln(x+5) as a logarithm of a single quantity

Answer 1

Kelly,
Hope you won't be annoyed, but I'm going to ask you to type out the three rules of logs that pertain to products, quotients and powers. Then we can jump in and solve this homework problem.

Answer 2

loga+logb=log (a*b)

loga-logb=log (a/b)

loga^x=x*loga

Answer 3

Great, Kelly. And you're not annoyed with me?
Now, using one of those three rules, re-write 2ln x.
Using another, re-write (1/2)ln (x+5).

Answer 4

Oh, not at all. :) You're nice and helping me. I like to thank you.

Answer 5

Actually, you need to use the same rule twice! Sorry about that typo.

Answer 6

Do i use loga-logb=log (a/b)?

Answer 7

Look over those three rules. Which one seems to apply to 2*ln x?

Answer 8

Note that there is neither multiplication nor division in 2*ln x.

Answer 9

Oh, umm...the third one?

Answer 10

Right. 2*ln x = ln (what?)

Answer 11

lnx*2

Answer 12

Great. Please try re-writing the other expression now, that is,
please re-write (1/2)ln (x+5) using the same rule as before.

Answer 13

lnx* 1/2=ln(x+5)*x? I think i messed up. ._.

Answer 14

You got the first one right the first time:\[2\ln x=\ln x ^{2}.\]
How? You moved the coefficient 2 from in front of the logarithm to the right of and above the argument of the ln function. Right on.

Do exactly the same thing with \[\frac{ 1 }{ 2 }\ln (x+5).\]

Answer 15

In the latter expression, the argument of the ln function is (x+5).

Answer 16

Using the methods that @mathmale has been talking about, the latter expression should be expressed as ln(x+5)*1/2

Answer 17

1/2ln(x+5)=ln(x+5)*1/2

Answer 18

Between us, fine. For a more fickle audience, it'd be better to enclose that 1/2 in parentheses for clarity: ln (x+5)^(1/2), and (more importantly) to use ^ for exponentiatin, not *. (*) signifies multiplication.

Answer 19

Thank you so much! sorry if I am slow. My teacher usually teach me log is a power and power is a log and etc.

Answer 20

to summarize:
\[2 \ln x - \frac{ 1 }{ 2 } \ln (x+5)=\ln x ^{2}-\ln (x+5)^{(1/2)}.\]

Answer 21

You're welcome, Kelly!
We have one more thing to do. Notice that (-) sign in the very last expression I typed?
Please go back to the 3 rules of logs and decide which one best fits this situation where you're subtracting one logarithm from anther. Want to try applying it?

Answer 22

Sure! and it is loga-logb=log(a/b)

Answer 23

Super. Now, can you apply that rule to our problem?

Answer 24

Sure. Few mins.

Answer 25

ln2^1/2

Answer 26

Here's what I see of what we've agreed upon so far:
\[\ln x ^{2} -\ln(x+5)^{(1/2)}\]
Think of this as ln a - ln b (similar to what you said 5 minutes ago).
ln a - ln b = ln (a/b) (which you have already given to me).

Then, similarly,

Answer 27

It was confusing DX

Answer 28

\[\ln x ^{2}-\ln(x+5)^{(1/2)}=\ln \frac{ x ^{2} }{(x+5)^{(1/2)}}\]

Answer 29

See the pattern we're following here? How much sense does this make to you?

Answer 30

You made it very clear. Thank you. I apologize if I am wasting your time

Answer 31

It's been very obvious that you want to learn this material, so there was absolutely no time wasted!

Answer 32

I do need to get off my computer (I've been doing this for something like 6 hours or more today), but hope to have the opp. to work with you again! Good night and best wishes to you.

Answer 33

Oh! thank you so much! Wow 6 hours? :o and have a nice day and goodnight!

Answer 34

Bye, Kelly! Same to you.

Answer 35

Bye mathmale! and thank you.