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

Mathematics
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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

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Other answers:

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? :) .
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)
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
@Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.
That was my next plan :P .
:)
@lucenzo do you know about the change of base formula for logs?
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 .
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.
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..
Oh, I've used that formula LOL. I actually was thinking that from the beginning. In this case I did \[\log_{a}n \] first
Well actually you don't have to use change of base :) .
Wait. Hold yon. You might :P .
Yeah you have too.\\
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: \[x^{2} + 6x - 9991 = 0\]
did you get that? @Dido525
I got x =-3/2, -1 haven't checked it however...
did you get this trinomial tho: x^2+6x−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
Use the properties of logs to expand that right log: @lucenzo
I think you have a typo there @Dido525 - they should all be base (x+3)
I do. I will fix.
\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)\]
Yep!
\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]
Are you there? @lucenzo
yeah. So we can expand all of the logs right?
Yep!
@Dido525 your last equation is wrong - you have forgotten the other log term
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: \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]
Do you notice anything special about the last log term? @lucenzo
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"
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?
\[\log_{5x+9} (x+3)^2 = \frac{ 2}{\log_{x+3} (5x+9) }\] \[\log_{x+3} (5x+9)(x+3) = 1 +\log_{x+3} (5x+9)\] \[\log_{x+3} (5x+9)+\frac{ 2}{\log_{x+3} (5x+9) }=3\]
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?
Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.
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 :) .
@Dido525 - are you sure of your last step?
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.
:)
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.
Yep :) .
do you understand this step @lucenzo ?
Yeah, I'm gonna try it
I only get one solution >.> .
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.
I only get 1 answer :( .
there are 3 answers altogether - @lucenzo have you solved the quadratic in u yet?
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:\[a=b^c\]
are you stuck @lucenzo ?
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)
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)
it might be better if you showed your steps here so that we can help you spot where you may have made a mistake
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:
not quite right @Dido525
The 1 is incorrect right?
yes
\[5x+9^{\log_{5x+9}(x+3) } = 5x+9^1\]
erm...
@lucenzo you are over complicating this.
\[x+3=5x+9\]
that's right, actually.
yes - that is correct - but you could have got there in one step. remember:\[\log_ba=c\implies a=b^c\]
\[x-5x=9-3\]
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
@asnaseer : I got so far: |dw:1351991548862:dw|
\[x+3=(5x+9)^{\frac{ 1 }{ 2 }}\]
|dw:1351991727301:dw|
LOL^
@Dido525 where did you derive that first equation from? @lucenzo that is correct - now square both sides
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 >.< .
@Dido525 please let @lucenzo solve this
\[x^2 + x = 0\]
perfect!
now factorise and solve
Sorry about that anaseer.
np :)
I am kinda into the problem XD .
I know some problems are so intriguing that we feel compelled to try and solve them ourselves - it is only natural. :)
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.
correct - and thanks for posting such an interesting problem @lucenzo :)
I agree! I am in uni and this was the toughest log I have done in ages :) .
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
yw :) and keep up the hunger for knowledge my friend! :)
Calculus is very fun! You will enjoy it a lot :) .
Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)
Derivatives hehe...

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