Question

simplify the logarithm log3(x+1) - log3(3x^2-3x-6)+ log3(x-2)

Answer 1

PLEASE HELP!

Answer 2

MEDAL TO ANYONE WHO ANSWERS

Answer 3

Just remember these rules and you should be fine:
\[\large \log_bM+\log_bN=\log_bMN\]\[\large\log_bM-\log_bN=\log_b\frac{M}{N}\]

Answer 4

So I need to use both of those equations in order to get the answer?

Answer 5

Well, use these 'properties'.
So, in general, if it's a positive logarithm, it goes in the numerator, and if it's a negative logarithm, it goes in the denominator. This entire expression? It's actually very simple :)
Just combine everything into one logarithm using those rules I posted...

Answer 6

Okay! Thank you so much! I hope this works.

Answer 7

You might want to stick around, though. At least tell me what you get when you combine them into one logarithm...

Answer 8

Will do

Answer 9

When setting it up should I have gotten Log3 (x+1)/(3x^2-3x-6) * log3 (x-2). ?

Answer 10

That's right :) Let me... uhh... straighten that out for you :)
\[\huge \log_3\frac{(x+1)(x-2)}{3x^2-3x-6}\]

Answer 11

Now, were it as simple as this, the problem wouldn't be half as awesome :)
Could you factor out that denominator, completely?

Answer 12

No because you get a gcf of 3 and you're left with 3(x^2-x-2)

Answer 13

By the way, you were only half-done with that...
\[\huge \log_3\frac{x+1}{3x^2-3x+6}+\log_3{(x-2)}\]
I hope you understand why I brought the (x-2) into the first logarithm. If you do, we'll proceed.

Answer 14

Isn't it because the log bases are the same?

Answer 15

Yeah, and the property, remember?
\[\large \log_bM + \log_bN =\log_bMN\]

Answer 16

Yep, got it!

Answer 17

So now I have Log3 (x^2-2x+1x-2) / 3(x^2-x-2)

Answer 18

So you said factor out the 3 in the denominator. Fair enough...
\[\huge \log_3\frac{(x+1)(x-2)}{3(x^2-x-2)}\]
But that's not done yet, you can still factor out that denominator.

Answer 19

I have no idea haha.

Answer 20

Wait no just kidding I do, it would be 3(x-2)(x+1) right,

Answer 21

Hehe... oh you of little faith :)
You're right, though, let's put that up...
\[\huge \log_3\frac{(x+1)(x-2)}{3(x-2)(x+1)}\]
And surely you must notice something :P

Answer 22

THEY CANCEL OUT :D

Answer 23

Yep, leaving you with...?

Answer 24

Log3 3

Answer 25

Whoops... a bit off, did you forget that 3 was in the denominator? :P

Answer 26

Oh crap log3 (1/3) ?

Answer 27

Very good. Now what is this equal to?

Answer 28

I have no idea

Answer 29

Well, what to what exponent do you raise 3 so that you get 1/3?

Answer 30

-1 ?

Answer 31

Bingo :P
So that long complicated logarithm, was, in the end, just -1

The End
XD

Answer 32

Oh god bless. I hate math, you are amazing. Would you like to help me with more?

Answer 33

Sure... hit me :D

Answer 34

Psychologists use an exponential model of the learning process, f(t)= c(1-e^-kt), where c is the total number of tasks to be learned, k is the rate of learning, t is the time, and f(t) is the number of tasks learned. Suppose you move to a new school and you want to learn the names of 25 classmates in your homeroom. If your learning rate for new tasks is 20% per day, how many complete names will you know after 2 days? After 8 days?

Answer 35

Sorry, Julia, I accidentally unplugged this laptop forgetting I removed the battery.
I'm gonna need time to digest this...

Answer 36

That's fine!

Answer 37

Okay, it seems like your garden-variety plugging in task.
You're given a function f(t) = c(1 - e^-kt)
And you're to find the value of this function. Well, it's all in the power of interpretation!
This would be a straightforward task if you know what c, k and t are, right?

Answer 38

So you would solve it twice, once for 2 and 8?

Answer 39

First of all, what are 2 and 8? Are they values for c? k? or t?

Answer 40

T because it's time right?

Answer 41

Very good :)
So t = 2 and t = 8.

What's c?

Answer 42

25 I think

Answer 43

I hope you're not guessing, but you're right. Your instincts must never fail you :)
Now what's k, then?

Answer 44

20% which would be .02

Answer 45

No....
20% is 0.2
You never said that ^
XD

Answer 46

Oh whoops haha

Answer 47

I got about 8 names

Answer 48

Oh yeah, it won't be exact, since there's an e... anyway, I'll leave the computational bit to your calculator. I have faith in both of you :)

Answer 49

Does 8 names and 20 names sound about right?

Answer 50

I think so? I haven't input it in any calculator :)

Answer 51

What is the inverse of log2 4x ?

Answer 52

Okay, so you have
y = log2 4x, right?
To get the inverse, switch y and x
x = log2 4y
And try to isolate y :)

Answer 53

Y= x-log2 /4 ?

Answer 54

No... remember that the inverse of the logarithm involves exponents.

Answer 55

Remember the definition of logarithm...
\[\huge a=\log_bM \iff b^a=M\]

Answer 56

In that case would a be y

Answer 57

Well, to make it simple, make it so that a is x and M is 4y, and b = 2
And you'll get your answer.

Answer 58

So y=2^x 4

Answer 59

You mean
\[\large \frac{2^x}{4}\]
right? :P

Answer 60

Oh.. That works too. Jesus I suck

Answer 61

And would the inverse of y=log3 X be y=(1/3)^x ?

Answer 62

Nope... try again :D

Answer 63

Y=3^x ?

Answer 64

Much better.

Answer 65

The value of an industrial machine has a decay factor of 0.75 per year. After six years, the machine is worth $7500. What was the original value of the machine?

Answer 66

I know you set it up with A=pe^rt but I'm not sure what to plug in since you're kind of working backwards with this one

Answer 67

I don't know how to do this o.O

Answer 68

Oh just kidding. What about (1/3)ln x + ln2- ln3 =3

Answer 69

Properties of logarithms, just include this
\[\huge p\log_bM=\log_bM^p\]

Answer 70

I've never seen this before. Do you have a dumber version haha?

Answer 71

This is as simple as it gets. In plain English, it says that if you multiply something to a logarithm, you can bring it up to become the exponent of that thing being, uhh, logaritmised :P

Answer 72

So what would go where with the equation I have?

Answer 73

With this...
\[\huge \frac13\ln x=\ln x^{\frac13}=\ln \sqrt[3]{x}\]

Answer 74

So is 3sqrt of x the answer?

Answer 75

No. Far from it, you have to solve for x
\[\huge \ln\sqrt[3]{x}+\ln2-\ln3=3\]

Answer 76

Sorry, but I have to go to bed now...
Us mere mortals need our sleep :)
Good luck :)
----------------------
Terence out