Question

I need help!!

Answer 1

\[\log_{6} (3x+2)=\log_6(x+4)+1\]

Answer 2

@LarsEighner can you help?

Answer 3

For a start, I'd put all the log terms on one side.

Answer 4

what I thought I was suppose to do since both sides have the same base

Answer 5

the we inverse it or something

Answer 6

What does log a - log b mean? What is it equal too?

Answer 7

log(a/b)?

Answer 8

Right. Now how to express 1 as a log of the same base?

Answer 9

wait what?

Answer 10

1 = log_6( ?)

In other words, convert the 1 to log_6

Answer 11

wouldnt it be 1=log_6(x)?

Answer 12

i have no clue

Answer 13

No. What is log_6(6) ?

Answer 14

what do you want me to do, isn't that just like true?
log_6(6) is true

Answer 15

im not sure what your asking here

Answer 16

Replace the 1 in your equation with log_6(something). You should know what something is.

Answer 17

\[ \large \log_{6} (3x+2)=\log_6(x+4)+1 \]

You did this:
\[ \large \log_{6} (3x+2) - \log_6(x+4) = 1 \]

And you knew you could replace the difference of the logs with

\[ \large \log_{6} {(3x+2\over (x+4)} = 1 \]

Now I am saying replace the 1 on the right with its value as a log base 6.

Answer 18

Actually what I did was this, (not sure if correct or not)
\[6^{\log(3x+2)}=6^{\log(x+4)+1}\] then
3x+2=(x+4)+1
3x+2=x+5
3x-x=5-2
2x=3
x=3/2

Answer 19

did i do it right?

Answer 20

wait why is it 22/7, when you are dividing the -3 not 7?

Answer 21

it would be 22/-3, at least thats what you are trying to say there

Answer 22

but that isn't the final answer right?

Answer 23

\begin{align} \large \log_{6} {{(3x+2)}\over {(x+4)}} &= 1 \cr \log_{6} {{(3x+2)}\over {(x+4)}} &= log_6 6 \cr {{(3x+2)}\over {(x+4)}} &= 6 \cr
(3x+2) &= 6 (x+4) \cr 3x+2 &= 6x+24\cr -3x &= 22 \cr x &= -{22 \over 3}\cr\end{align}

Answer 24

and thats it?

Answer 25

cause if so, i'm not understanding how you got the log_6 (3x+2)/(x+4)=1

Answer 26

Do you know anyone else that might help explain this to me?

Answer 27

@LarsEighner

Answer 28

We got to that step because log a - log b = log (a/b) for any base

Answer 29

@SolomonZelman I think I went wrong here somehow. Would you look it over?

Answer 30

@mathstudent55 ?

Answer 31

it uh says they are offline.. @LarsEighner

Answer 32

\[\large \log_{6} (3x+2)=\log_6(x+4)+1\]

\[\large \log_{6} (3x+2)-\log_6(x+4) =1\]

\[\large \log_{6} {{3x+2} \over {x+4}} = log_6 6\]

\[\large {{3x+2} \over {x+4}} = 6\]

\[\large 3x+2 =6 (x+4) \]

\[\large 3x+2 =6x+24 \]

\[\large 3x=6x+22) \]

\[\large -3x=22 \]

What's bothering me is it is negative.

\[\large x=- {22 \over 3}\]

Answer 33

@thomaster ?