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\[\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4\]
Is that the question?

yeah, that's it

They don't have the same base?

Darn... This complicates things...

Okay I would factor.

\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))\]

=4

Can you see anyway to simplify this? :) .

no, I have no idea

You have two logs multiplied. What can you do? :) .

They don't have the same base so you can't do "loga + logb = log(a*b)"

I meant for the right log :) .

OH

Ohh you need help factoring? Okay:
|dw:1351987084250:dw|

You do log (5x^2 + 24x + 27)/(x+3)

Wait I am still in the process of solving this :P .

and do that for both, lol okay. I'll try solving it here like that

Wow this is the hardest log I have done so far :P .

yeah, same. Never seen anything like it. But I think it'll work out nicely

That was my next plan :P .

:)

no. I think I might have to solve this question w/o using the formula, since we never learned it

I can't see how to solve this without using that formula?

okay, nvm then. Can you tell me what the formula is?

But you have too O_o .

converting bases isn't advanced... iirc.

change of base is given by:\[\log_ax=\frac{\log_bx}{\log_ba}\]

you usually learn it first...

that is what thought @Algebraic!

*what I thought

hm..

Well actually you don't have to use change of base :) .

Wait. Hold yon. You might :P .

Yeah you have too.\\

I got x =-3/2, -1
haven't checked it however...

did you get this trinomial tho: x^2+6xâˆ’9991=0

I got it!!!

No you don't get that...

I think I did it wrong

lol okay, yeah

@Algebraic! I got only -3/2 .

\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3)) \]
We were at that step.

I get 3 solutions

I do. I will fix.

\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)\]

Yep!

\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

yeah. So we can expand all of the logs right?

Yep!

I am ignoring that one for now :P .

I didn't forget it.

then it shouldn't be set "= 4"

Yeah, my mistake.

it would be 3 then? Instead of 4

So expanding all that you should get:
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

Good!

So you should get :
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3\]

I got to that part no problem. The next part is what confuses me xD

Now what do you notice about the first log term?

yeah, I expanded the first term

No no!!

Remember, in a log you can bring that power down :) .

|dw:1351988960498:dw|

powering every term in the equation by the same number?

quadratic in
\[\log_{x+3} (5x+9) \]

ah, that

I see now xD
thanks you

u^2 -3u+2=0

So you should get :
\[2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3\]

yes, I did

Now use change of base that you learned a little earlier. :) .

|dw:1351989182904:dw|

I know the normal way of using that rule where it would be: \[\log(x+3)/\log(5x+9) \]

Hint: Use it on the right log :) .

are we assuming that the "y" is just 10. so its log base of 10?

Well in this case it's base x+3 .

no, i meant for the first term. the term on the left

It's base 5x+9 .

just for the \[\log_{5x+9}(x+3) \]

The base is 5x+9 for that yes.

yeah, that's why i did \[\log(x+3)/\log(5x+9)\]

The trick for the right log is that you want to convert that right base to the base on the left log.

That's correct but you want base 5x+3 .

5x+9*

Yeah. My bad.

I got 1/\[1/\log_{5x+9}(x+3) \]

Great job!

whoops, just meant to put 1 "1/"

and now we put everything to the power of *5x+9* ?

So you should now have:
\[2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)\]

Move the common terms to one side.

nvm

yeah lol, I was about to say that

It's fine :) .

Yes.

remember you started from:\[2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9} (x+3)} =3\]

Ohh!!!!

can we not put it all to ther power of 5x+9 tho? I'm still thinking it might be simpler

oops.

:)

Yep :) .

Yeah, I'm gonna try it

I only get one solution >.> .

I only get 1 answer :( .

yeah I got u = 1 or 1/2

so 1 or 0.5

what would be the third answer?

that is correct. so now you need to solve:\[log_{5x+9}(x+3)=1\]and:\[log_{5x+9}(x+3)=0.5\]

right

so, starting with:\[log_{5x+9}(x+3)=1\]what do you get for x?

Now I got it at last :D .

:)

remember that if:\[\log_ba=c\]then:\[a=b^c\]

yeah, but I just realized I had a 2 multiplied to the log term (I'm gonna try it without the 2)

I got 3/2 for the first one (if x = 1)

okay

I got -3/2 , 0 , 1.

I took the equation you had then I put both sides to the power of 5x+9 like so:

The 1 is incorrect right?

yes

\[5x+9^{\log_{5x+9}(x+3) } = 5x+9^1\]

erm...

\[x+3=5x+9\]

that's right, actually.

\[x-5x=9-3\]

yeah it's correct but why would you do that? :( .

That's the way we learned to write it

same thing...

Just a way of showing how to get there, it's just the same thing

ok - fair enough - if that is how you were taught then you should stick to that method.

I think I just showed the work

Yeah it's correct regardless.

yep you did.

ok so what x value do you get from this?

3/2

6/4 which reduces to 3/2

I keep getting - 3/2 hmm...

this step was correct:\[x-5x=9-3\]which then leads to:\[-4x=6\]

uh-oh, I made a sign error xD. I wrote -6

you're right. It's -3/2

ok, so first solution is x=-1.5.
next, you need to solve this for x:\[log_{5x+9}(x+3)=0.5\]

again here it would help if you listed your steps

okay

\[x+3=(5x+9)^{\frac{ 1 }{ 2 }}\]

|dw:1351991727301:dw|

LOL^

You get 0 and 1 if you solve for x.

If you solve x I keep getting 0 and 1.

This thread is so epic I am lagging >.< .

\[x^2 + x = 0\]

perfect!

now factorise and solve

Sorry about that anaseer.

np :)

I am kinda into the problem XD .

i did the quadratic formula and got x = -1 or 0

you have the right answer

but you did not need to use the quadratic formula

Yeah I got that too :) . I just had a silly algebra error.

You could have factored :P .

wha

\[x^2+x=0\]\[x(x+1)=0\]so\[x=0\]or:\[x=-1\]

-.-

so final solution is:
x = -1.5 or -1 or 0

yeah, and I think ALL of them work. so none are extraneous

or -3/2, -1, 0 .

Yeah. I checked. All work.

I agree! I am in uni and this was the toughest log I have done in ages :) .

yw :)
and keep up the hunger for knowledge my friend! :)

Calculus is very fun! You will enjoy it a lot :) .

Derivatives hehe...