Question

Evaluate each expression: Explain how to work it out. *Problems will be in comment box shortly*

Answer 1

@matt101 please help me

Answer 2

Which one do you need help with?

Answer 3

@matt101 I need help on the first one to help get me going, but 42 I need help on as well; 44 I said 2x, but I am not for sure if it's right?

Answer 4

@perl

Answer 5

Alright well starting with 41, what are your steps?

And for 43, -3 is correct!

Answer 6

okay thank you @matt101 see I do my work.

Answer 7

@matt101 now I need to get to bed, due to no one helping me on here on my question before you helped me out @matt101

I don't know what the 1st step to #41 is

Answer 8

No problem! Shoot me a message tomorrow when you want to work through these and I'll give you a hand.

Answer 9

ok which one would you like to work on

Answer 10

#41 please quick, because I shouldn't be up past 12:00AM and it's 12:23AM

Answer 11

$$ \Large {
\log_{11} 11(n-5)

}


$$

Answer 12

what are the directions

Answer 13

it moves to the right 5 units

Answer 14

$$ \large {
\log_{11} 11(n-5)= \log_{11} 11 +\log_{11}(n-5) = 1 + \log_{11}(n-5)

}


$$

Answer 15

May I ask how did you jump to that conclusion of 1 + log11 (n - 5)

Answer 16

I used the rule
Log ( A*B ) = Log(A) + Log(B)

Answer 17

see I didn't even know about that.

Answer 18

what's the next step

Answer 19

i dont think we can simplify that further

Answer 20

is the answer 1 + log11 (n - 5)

Answer 21

what were the directions?

Answer 22

is it up 1 unit and to the right 5 units

Answer 23

can you upload the directions sheet.

Answer 24

direction sheet for what

Answer 25

what did you want to do with this log. What was the original question

Answer 26

evaluate each expression

Answer 27

there is no value for n given

Answer 28

correct?

Answer 29

so what's the evaluated expression for 41. log 11 (n - 5) + 1

Answer 30

you can't simplify it further, and can't evaluate it either

Answer 31

thats why i was confused by the directions

Answer 32

it's just evaluate each expression

Answer 33

okay I am making sure that's the answer

Answer 34

is 44 2x

Answer 35

2 + log x

Answer 36

$$\Huge {2 + \log_4 (x)} $$

Answer 37

oh well so is log4 (x) + 2 the same as what you said

Answer 38

yes :)

Answer 39

My question is how to solve #42, becauase it's a lot different than the other ones

Answer 40

that cant be simplified further

Answer 41

really are you sure @perl 6log6 (3x + 2) can't be evaluated any further

Answer 42

\(Log (a*b)=Log(a)+log(b)\)
\(Log(a/b)=log(a)-log(b)\)
\(alogb=log(b)^a\)

These are the rules that you you use to evaluate any logs.

Answer 43

Here, you have \(6log_6(3x+2)\)
There's no rule for \(log a+b\) that means that it cant be evaluated further