Question

Use properties of logarithms to expand the logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator.

ln [ (x^10)sqrt((x^2)+8)/(x+8)^3)]

Answer 1

Nobody is helping me. Please help me @SolomonZelman PLEASE!

Answer 2

\[\Large\rm \ln\left[\frac{x^{10}\sqrt{x^2+8}}{\color{orangered}{(x+8)^3}}\right]\]Let's start with this denominator first.
We'll apply a rule of logs:\[\Large\rm \log\left(\frac{a}{\color{orangered}{b}}\right)=\log(a)-\log(\color{orangered}{b})\]

Answer 3

GOD THANK YOU

Answer 4

Understand how we can make use of this rule? :o

Answer 5

lol

Answer 6

so log(a)-log(x+8)^3

Answer 7

?

Answer 8

\[\Large\rm \ln\left[\frac{x^{10}\sqrt{x^2+8}}{\color{orangered}{(x+8)^3}}\right]=\ln\left[x^{10}\sqrt{x^2+8}\right]-\ln\left[\color{orangered}{(x+8)^3}\right]\]Mmmm ok good, that's our first step.

Answer 9

This is gonna be like uhhhh... 8 or 9 step problem -_-
so you better get comfy...

Answer 10

oh i am trust me

Answer 11

Similarly we can break up the product in the first term using another rule of logs.\[\Large\rm \log(a\cdot \color{royalblue}{b})=\log(a)+\log(\color{royalblue}{b})\]

Answer 12

See how we can use that maybe? :d
\[\Large\rm \ln\left[x^{10}\color{royalblue}{\sqrt{x^2+8}}\right]-\ln\left[(x+8)^3\right]\]

Answer 13

ok ok

Answer 14

So that step takes us to,\[\Large\rm \ln\left[x^{10}\right]+\ln\left[\color{royalblue}{\sqrt{x^2+8}}\right]-\ln\left[(x+8)^3\right]\]

Answer 15

Another rule of logs will let us write our exponents out in front of the log as a coefficient,\[\LARGE\rm \log\left(a^{\color{#CC0033}{b}}\right)=\color{#CC0033}{b}~\log(a)\]

Answer 16

Let's apply it to this last term first,
\[\LARGE\rm \ln\left[(x+8)^{\color{#CC0033}{3}}\right]\]

Answer 17

What do you get for that term after applying the rule? c:

Answer 18

3log(x+8)

Answer 19

\[\LARGE\rm \color{#CC0033}{3}\ln\left[(x+8)\right]\]Mmmm k good :)

Answer 20

right!!

Answer 21

Here is where we're at currently,\[\Large\rm \ln\left[x^{10}\right]+\ln\left[\sqrt{x^2+8}\right]-3\ln\left[x+8\right]\]

Answer 22

So what is the next step? Sorta following

Answer 23

You're doing a great job

Answer 24

lol you too c:
Let's apply that same rule to the `first term`.
What does it give you?\[\Large\rm \ln\left[x^{10}\right]=?\]

Answer 25

10ln(x)?

Answer 26

\[\Large\rm 10\ln\left[x\right]+\ln\left[\sqrt{x^2+8}\right]-3\ln\left[x+8\right]\]Ok great.
We also want to apply this same rule to the `middle term`.
But we have to mess with it before we can do so.

Answer 27

Recall:\[\Large\rm \sqrt{x}=x^{1/2}\]We can write the square root like that so that we can apply the rule easier.

Answer 28

\[\Large\rm 10\ln\left[x\right]+\ln\left[\sqrt{x^2+8}\right]-3\ln\left[x+8\right]\]
\[\Large\rm 10\ln\left[x\right]+\ln\left[(x^2+8)^{1/2}\right]-3\ln\left[x+8\right]\]Understand what I did there?

Answer 29

Woah I dunno why it's in bold like that +_+

Answer 30

yes! :) So that would be 1/2ln(x^2+8) ?

Answer 31

\[\Large\rm 10\ln\left[x\right]+\frac{1}{2}\ln\left[x^2+8\right]-3\ln\left[x+8\right]\]Mmmm very good.
And believe it or not, that is our final answer!! \c:/
It's finally over! hehe

Answer 32

seriously thank you so much!!!!!!!!!!! thank you for answering me!!!!! I have a couple more questions do you think you can possibly help me with those too :))

Answer 33

Homework due tonight at 10 I've been working on these problems all day haha

Answer 34

Open a new thread with your other questions c:

I'm really tired but I'll try to stop by ^^ heh

Answer 35

Okay I'll do that!

Answer 36

I do math tutoring for my job, so answering math questions all day long can be draining XD lol

Answer 37

Yep makes sense I totally understand!!!