Question

is log(8x)-log(x^2-1)=log(3) be the same as
8x-(x^2-1)=3

Answer 1

no it is not

Answer 2

you need to combine the logs on the left side using this formula

log(A) - log(B) = log(A/B)

Answer 3

(sorry I don't think it will make a difference but all the logs are base 2)

Answer 4

no that doesn't affect the answer

Answer 5

ok... so what would this be in this case? would that rule apply if it was log()+log()?

Answer 6

if you add 2 logs of the same base, you use log(A)+log(B) = log(A*B)

Answer 7

sorry to ask this, but in what case would that rule apply then?

Answer 8

which rule

Answer 9

log() log()=log()
is the same as
() ()=()
(I thought the symbols inside were + or -)

Answer 10

I originally thought it was log()+log()=log()
is the same as
()+()=()

Answer 11

no the log rules are given here

http://www.purplemath.com/modules/logrules.htm

Answer 12

it's handy to have them memorized

Answer 13

once you have log(A) = log(B), you can say A = B

Answer 14

Ah I see.

Answer 15

that's why you combine the logs on the left side

to have log = log

Answer 16

so I don't have to redo
log(-3x-1)=log(-4x-6)?

Answer 17

no because that's in the form log(A) = log(B)

A = -3x-1
B = -4x-6

Answer 18

ah I see.
so... sorry to ask this, but can you help me through with these types of problems?
I have
log(8x/x^2-1)=log(3)

Answer 19

so you can now say

\[\Large \frac{8x}{x^2-1} = 3\]

Answer 20

would the x and x^2 cancel so it becomes
log(8/x-1)?

Answer 21

no you can't do that

Answer 22

Ok
sorry I didn't notice the log() removed.

Answer 23

also, you can't cancel the x's like that either

Answer 24

multiply both sides by x^2-1 to get 8x=3x^2-3?

Answer 25

good

Answer 26

get everything to one side, then use the quadratic formula

Answer 27

you could factor, but factoring doesn't always work

Answer 28

so then zero out one side
3x^2-8x-3=0
then factor it?
(3x+1)(x-3)

Answer 29

that works too

Answer 30

then, the options for an answer would be
0, -1/3, and 3?

Answer 31

just need to check which one works and take out the option for 0?

Answer 32

how are you getting 0?

Answer 33

sorry IDK must've been a mistake.
but, it should be
log(8(3))-log(9-1)=log(3)
log(24)-log(8)=log(3)
log(24/8)=log(3)
24/8=3

Answer 34

so 3 is definitely a solution, check -1/3 as well

Answer 35

sorry it took a while

Answer 36

its fine, take all the time you need

Answer 37

log(8(-1/3))-log(-1/3)^2-1)=log(3)
log(-8/3)-log(1/9-1)=log(3)
log(-8/3/1/9-1)=log(3)
(-8/3)/(1/9)-1=log(3)
(-8/3)/(-8/9)=3
3=3
so both 3 and -1/3 are answers

Answer 38

did I do it right?

Answer 39

you nailed it

Answer 40

Thank you!
my mind was blown when I realized how simple it was!
Thank you thank you thank you.

Answer 41

you're welcome

Answer 42

I have one more question... I need someone to check it to make sure it's correct

Answer 43

3^x=17
solve for the variable

Answer 44

this won't get ore simplified without going into decimals than
x=log(17)/log(3) right?

Answer 45

*more not ore sorry

Answer 46

yeah basically \[\Large x = \log_{3}(17)\] and you can use the change of base formula to say

\[\Large x = \log_{3}(17) = \frac{\log(17)}{\log(3)}\]

Answer 47

Ok. thanks again. I really need to get myself together for this HW.

Answer 48

you're doing quite well actually

Answer 49

thanks... :P
you're a really great helper too. :)
I think I can cover the rest of these now that I got what you meant.
thank you so much.

Answer 50

you're welcome