I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4

- anonymous

I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4

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- anonymous

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

- anonymous

@lucenzo

- anonymous

yeah, that's it

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## More answers

- anonymous

They don't have the same base?

- anonymous

Darn... This complicates things...

- anonymous

Okay I would factor.

- anonymous

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

- anonymous

=4

- anonymous

Can you see anyway to simplify this? :) .

- anonymous

no, I have no idea

- anonymous

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

- anonymous

This is a type of a bonus, so we never learned this (I thought of factoring but idk how to xD). Is it possible to get both logs to the same base? By putting them both to the power of (5x+9)(x+3)

- anonymous

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

- anonymous

I meant for the right log :) .

- anonymous

OH

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- asnaseer

@Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.

- anonymous

That was my next plan :P .

- asnaseer

:)

- asnaseer

@lucenzo do you know about the change of base formula for logs?

- anonymous

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

- asnaseer

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

- anonymous

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

- anonymous

But you have too O_o .

- asnaseer

I would assume that if this is a /bonus/ question, then the teacher expected some of you to do some learning on your own to try and solve this.

- anonymous

converting bases isn't advanced... iirc.

- asnaseer

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

- anonymous

you usually learn it first...

- asnaseer

that is what thought @Algebraic!

- asnaseer

*what I thought

- anonymous

hm..

- anonymous

Oh, I've used that formula LOL. I actually was thinking that from the beginning. In this case I did \[\log_{a}n \] first

- anonymous

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

- anonymous

Wait. Hold yon. You might :P .

- anonymous

Yeah you have too.\\

- anonymous

then I faactored both trinomials and it was log(base 10) so I was able to multiply the logs and now I've ended up w/ a strange trinomial: \[x^{2} + 6x - 9991 = 0\]

- anonymous

did you get that? @Dido525

- anonymous

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

- anonymous

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

- anonymous

I got it!!!

- anonymous

No you don't get that...

- anonymous

I think I did it wrong

- anonymous

lol okay, yeah

- anonymous

@Algebraic! I got only -3/2 .

- anonymous

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

- asnaseer

I get 3 solutions

- anonymous

Use the properties of logs to expand that right log: @lucenzo

- asnaseer

I think you have a typo there @Dido525 - they should all be base (x+3)

- anonymous

I do. I will fix.

- asnaseer

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

- anonymous

Yep!

- anonymous

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

- anonymous

Are you there? @lucenzo

- anonymous

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

- anonymous

Yep!

- asnaseer

@Dido525 your last equation is wrong - you have forgotten the other log term

- anonymous

I am ignoring that one for now :P .

- anonymous

I didn't forget it.

- asnaseer

then it shouldn't be set "= 4"

- anonymous

Yeah, my mistake.

- anonymous

it would be 3 then? Instead of 4

- anonymous

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

- anonymous

@lucenzo

- anonymous

Do you notice anything special about the last log term? @lucenzo

- anonymous

yeah, its the same thing over the same thing. So it is 1. You move that to the other side and the "4" becomes a "3"

- anonymous

Good!

- anonymous

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

- anonymous

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

- anonymous

Now what do you notice about the first log term?

- anonymous

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

- anonymous

yeah, I expanded the first term

- anonymous

No no!!

- anonymous

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

- anonymous

|dw:1351988960498:dw|

- anonymous

powering every term in the equation by the same number?

- anonymous

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

- anonymous

ah, that

- anonymous

I see now xD
thanks you

- anonymous

u^2 -3u+2=0

- anonymous

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

- anonymous

yes, I did

- anonymous

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

- anonymous

|dw:1351989182904:dw|

- anonymous

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

- anonymous

Hint: Use it on the right log :) .

- anonymous

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

- anonymous

Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.

- anonymous

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

- anonymous

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

- anonymous

It's base 5x+9 .

- anonymous

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

- anonymous

The base is 5x+9 for that yes.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

5x+9*

- anonymous

Yeah. My bad.

- anonymous

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

- anonymous

Great job!

- anonymous

whoops, just meant to put 1 "1/"

- anonymous

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

- anonymous

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

- anonymous

Move the common terms to one side.

- anonymous

nvm

- anonymous

yeah lol, I was about to say that

- anonymous

It's fine :) .

- asnaseer

@Dido525 - are you sure of your last step?

- anonymous

Yes.

- asnaseer

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

- anonymous

Ohh!!!!

- anonymous

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

- anonymous

oops.

- asnaseer

:)

- asnaseer

from this point, it might be better to do what @Algebraic! did, i.e. let:\[u=log_{5x+9}(x+3)\]to get:\[2u+\frac{1}{u}=3\]and then solve this quadratic in u first.

- anonymous

Yep :) .

- asnaseer

do you understand this step @lucenzo ?

- anonymous

Yeah, I'm gonna try it

- anonymous

I only get one solution >.> .

- asnaseer

we got to:\[2u+\frac{1}{u}=3\]multiply both sides by u to get:\[2u^2+1=3u\]\[2u^2-3u+1=0\]this should have 2 solutions for u.
then use the fact that:\[u=log_{5x+9}(x+3)\]to find all the x values.

- anonymous

I only get 1 answer :( .

- asnaseer

there are 3 answers altogether - @lucenzo have you solved the quadratic in u yet?

- anonymous

yeah I got u = 1 or 1/2

- anonymous

so 1 or 0.5

- anonymous

what would be the third answer?

- asnaseer

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

- anonymous

right

- asnaseer

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

- anonymous

Now I got it at last :D .

- asnaseer

:)

- asnaseer

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

- asnaseer

are you stuck @lucenzo ?

- anonymous

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

- anonymous

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

- asnaseer

plz double check your work - I think you have made an error in the "sign" of the result. and it is for (u=1) not (x=1)

- asnaseer

it might be better if you showed your steps here so that we can help you spot where you may have made a mistake

- anonymous

okay

- anonymous

I got -3/2 , 0 , 1.

- anonymous

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

- asnaseer

not quite right @Dido525

- anonymous

The 1 is incorrect right?

- asnaseer

yes

- anonymous

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

- anonymous

erm...

- asnaseer

@lucenzo you are over complicating this.

- anonymous

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

- anonymous

that's right, actually.

- asnaseer

yes - that is correct - but you could have got there in one step. remember:\[\log_ba=c\implies a=b^c\]

- anonymous

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

- anonymous

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

- anonymous

That's the way we learned to write it

- anonymous

same thing...

- anonymous

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

- asnaseer

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

- anonymous

I think I just showed the work

- anonymous

Yeah it's correct regardless.

- anonymous

yep you did.

- asnaseer

ok so what x value do you get from this?

- anonymous

3/2

- anonymous

6/4 which reduces to 3/2

- anonymous

I keep getting - 3/2 hmm...

- asnaseer

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

- anonymous

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

- anonymous

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

- asnaseer

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

- asnaseer

again here it would help if you listed your steps

- anonymous

okay

- anonymous

@asnaseer :
I got so far:
|dw:1351991548862:dw|

- anonymous

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

- anonymous

|dw:1351991727301:dw|

- anonymous

LOL^

- asnaseer

@Dido525 where did you derive that first equation from?
@lucenzo that is correct - now square both sides

- anonymous

You get 0 and 1 if you solve for x.

- anonymous

If you solve x I keep getting 0 and 1.

- anonymous

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

- asnaseer

@Dido525 please let @lucenzo solve this

- anonymous

\[x^2 + x = 0\]

- asnaseer

perfect!

- asnaseer

now factorise and solve

- anonymous

Sorry about that anaseer.

- asnaseer

np :)

- anonymous

I am kinda into the problem XD .

- asnaseer

I know some problems are so intriguing that we feel compelled to try and solve them ourselves - it is only natural. :)

- anonymous

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

- asnaseer

you have the right answer

- asnaseer

but you did not need to use the quadratic formula

- anonymous

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

- anonymous

You could have factored :P .

- anonymous

wha

- asnaseer

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

- anonymous

^

- anonymous

-.-

- asnaseer

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

- anonymous

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

- anonymous

or -3/2, -1, 0 .

- anonymous

Yeah. I checked. All work.

- asnaseer

correct - and thanks for posting such an interesting problem @lucenzo :)

- anonymous

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

- anonymous

lol. I'm the one to thank you guys. Thank you sooo much! You all really helped my very much. I'm in high school right now, and when I take calculus next semester and have to do a log question like this, I'll feel like I'm ahead of the others xD

- asnaseer

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

- anonymous

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

- anonymous

Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)

- anonymous

Derivatives hehe...

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