Question

log[12]3+log[12]4

Answer 1

use property of logs to rewrite:

log[b]x+log[b]y=log[b]xy

Answer 2

then you can solve, evaluate

Answer 3

log[12]12

Answer 4

yep, which is....?

Answer 5

1....

Answer 6

you got it!!!

Answer 7

:) ok i have another

Answer 8

\[\log[4]\left( 2x ^{2} -x-1\right)-\log[4]\left( x-1 \right)+\log[4]16=4\]

Answer 9

i dont even know where to start with this

Answer 10

use the properties of logs to simplify...

log[b]x-log[b]y=log[b](x/y)

you can start by simplifying the one without x's in the argument

Answer 11

log[4]16x-5 for the end 2 ?

Answer 12

log[4](2x+1)=2 => 16 = 2x+1 => x = ?

Answer 13

how'd you get that?

Answer 14

log[4](2x^2-x-1)-log[4](x-1) = log[4](2x^2-x-1/x-1) = log[4](2x+1)

log[4]16=2

=>log[4](2x+1)+2=4 =>log[4](2x+1)=2 => 2x+1 = 4^2 solve for x

Answer 15

2x^2-x-1 = (2x+1)(x-1)

Answer 16

i dont understand :(

Answer 17

log[b]x-log[b]y = log[b]x/y do you understand that one?

Answer 18

yes

Answer 19

okay so log[4](2x^2-x-1) - log[4](x-1) = log[4](2x^2-x-1/x-1)
= log[4]((2x+1)*(x-1)/(x-1)) = log[4](2x+1)

do you follow that?

Answer 20

kinda

Answer 21

what are you shaky on?

Answer 22

(2x+1)*(x-1)/(x-1) where did you get all that

Answer 23

i factored 2x^2-x-1 into (2x+1)(x-1). the (x-1) in the denominator came from -log[4](x-1)

Answer 24

ok

Answer 25

you follow how log[4]16 = 2?

Answer 26

no

Answer 27

first, a log is just an exponent. it's the exponent you raise the base to in order to get the argument.

in log[4]16 4 is the base and 16 is the argument. so what power do you raise 4 to inorder to get 16?

Answer 28

2

Answer 29

yep so log[4]16 = 2... make sense?

Answer 30

but how did you get that from log[4]((2x+1)*(x-1)/(x-1))

Answer 31

hold on... did those separately and now we can bring together...

(log[4](2x^2-x-1)-log[4](x-1)) + log[4]16 = 4

=> log[4]((2x^2-x-1)/(x-1)) + log[4]16 = 4
=> log[4](2x+1)(x-1)/(x-1) + log[4]16 = 4
=> log[4](2x+1) + log[4]16 = 4
=>log[4](2x+1) + 2 = 4
=>log[4](2x+1) = 2
=>4^2 = 2x+1

then you can solve for x

do you follow this? if not, where are you getting goofed up...

Answer 32

ohhhh

Answer 33

?

Answer 34

i get it its just really big and confusing because i was gone for a week when they learned this and now im having a test on it tmr so im trying to understand it all

Answer 35

if you want to learn efficiently, spend a little time every day on it. no matter what it is. you have the tech to stay connected so you should never be behind, right? keep at it!

Answer 36

\[\frac{ 15 }{ 2 }=x \]

Answer 37

there you go... you should be able to plug into the original and check... if you want

Answer 38

thank you so much :)

Answer 39

you're welcome!