Question

help

Answer 1

\[\log_{3}(x-1)-\log_{3}(3x^2-3x-6)+\log_{3}(x-2)\]

Answer 2

its (x+1) not subtract

Answer 3

\[3^{\log _{3}(x) } = x\]

Answer 4

Can you explain it to me step by step?

Answer 5

Recall: log rules

log(a) - log(b) = log (a/b)
log(a) + log(b) = log(a*b)

Answer 6

i don't get it

Answer 7

Log(some quantity) - log(some other quantity) = log( the quotient of the two quantities)

Answer 8

log(x-1) - log (3x^2-3x-6) = \[\log _{3}(\frac{ x-1 }{ 3x^2 -3x - 6})\]

Answer 9

Its suppose to be x+1 my mistake

Answer 10

ok, then for the one that is added, you multiply instead of divide

\[\log _{3}[\frac{ x+1 }{ 3x^2 - 3x - 6 }*(x-2)]\]

Answer 11

Ok, what is the next step?

Answer 12

maybe factor the bottom term 3(x^2-x-2) = ...

Answer 13

3(x-2)(x+1)

Answer 14

\[\log _{3}[\frac{ (x+1)(x-2) }{ 3(x+1)(x-2) }]\]

Answer 15

right?

Answer 16

Im not sure.. I'm just starting to learn this stuff. So I'm taking your word for it… you understand this right?

Answer 17

yes, so you can cancel out both those quantities and be left with log base 3 of 1/3

Answer 18

how would i write that?

Answer 19

\[\log \frac{ 3 }{ 1/3}\]

Answer 20

\[\log_{3}1/3\]

Answer 21

\[\log _{3}[\frac{ 1 }{ 3 }]\]

Answer 22

yea

Answer 23

and thats it?

Answer 24

yeah

Answer 25

Thanks! Would you be able to help with one more problem?

Answer 26

sure

Answer 27

\[\log_{7}(2x-3)^2=2\]

Answer 28

For the last one real fast....

Answer 29

log base 3 of 1/3 also can be thought of as "3 to what power is 1/3"

Answer 30

3^(-1) = 1/3

Answer 31

so log base 3 of 1/3 = -1

Answer 32

Got it :)

Answer 33

For the next one, recall this rule for logs...

\[\log(x^2) = 2*\log x\]

Answer 34

pull the power down to the front

Answer 35

7log(2x-3)^2=2

Answer 36

?

Answer 37

\[2 *\log _{7}(2x-3) = 2\]

Answer 38

Ahhh i see ..

Answer 39

now divide both sides by 2

Answer 40

\[\log _{7}(2x-3) = 1\]

Answer 41

now recall this rule:

\[b ^{\log _{b}(x)} = x\]

Answer 42

7 raised to the log base 7 of x = x

Answer 43

so...

Answer 44

\[7^{\log _{7}(2x-3)} = 2x-3\]

Answer 45

see that?

Answer 46

yes and then 2x-3 on both sides cancel out?

Answer 47

no, that was just to show what the one side does, you have to do the same thing to both sides ....

Answer 48

\[7^{\log _{7}(2x-3)} = 7^{1}\]

Answer 49

2x - 3 = 7^1

Answer 50

2x - 3 = 7

Answer 51

ok..

Answer 52

then what do i do next?

Answer 53

2x = 10
x=5

Answer 54

That's it?

Answer 55

yeah, solve for x.. to check , let x=5

Answer 56

\[\log _{7}(2(5)-3) = \log _{7}(7)\]

Answer 57

7^(what power) = 7

(what power) = 1

Answer 58

the equation = 1 on the right side, so it is true, and x=5 is the solution

Answer 59

Got it! thanks so much :)

Answer 60

no prob, just memorize the 4 or 5 log rules that's all

Answer 61

I'll try :) We just started learning this so it's very new to me

Answer 62

Logs always cause trouble for some reason. Just remember all they are, are exponents

Answer 63

\[\log _{b}(x) = y\]

\[b ^{y} = x\]

Answer 64

log base b of x .... means.... b to what power = x,

Answer 65

b to the y power is x