anonymous
  • anonymous
I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4\] Is that the question?
anonymous
  • anonymous
@lucenzo
anonymous
  • anonymous
yeah, that's it

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anonymous
  • anonymous
They don't have the same base?
anonymous
  • anonymous
Darn... This complicates things...
anonymous
  • anonymous
Okay I would factor.
anonymous
  • anonymous
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))\]
anonymous
  • anonymous
=4
anonymous
  • anonymous
Can you see anyway to simplify this? :) .
anonymous
  • anonymous
no, I have no idea
anonymous
  • anonymous
You have two logs multiplied. What can you do? :) .
anonymous
  • 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
  • anonymous
They don't have the same base so you can't do "loga + logb = log(a*b)"
anonymous
  • anonymous
I meant for the right log :) .
anonymous
  • anonymous
OH
anonymous
  • anonymous
Ohh you need help factoring? Okay: |dw:1351987084250:dw|
anonymous
  • anonymous
You do log (5x^2 + 24x + 27)/(x+3)
anonymous
  • anonymous
Wait I am still in the process of solving this :P .
anonymous
  • anonymous
and do that for both, lol okay. I'll try solving it here like that
anonymous
  • anonymous
Wow this is the hardest log I have done so far :P .
anonymous
  • anonymous
yeah, same. Never seen anything like it. But I think it'll work out nicely
asnaseer
  • asnaseer
@Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.
anonymous
  • anonymous
That was my next plan :P .
asnaseer
  • asnaseer
:)
asnaseer
  • asnaseer
@lucenzo do you know about the change of base formula for logs?
anonymous
  • anonymous
no. I think I might have to solve this question w/o using the formula, since we never learned it
asnaseer
  • asnaseer
I can't see how to solve this without using that formula?
anonymous
  • anonymous
okay, nvm then. Can you tell me what the formula is?
anonymous
  • anonymous
But you have too O_o .
asnaseer
  • 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
  • anonymous
converting bases isn't advanced... iirc.
asnaseer
  • asnaseer
change of base is given by:\[\log_ax=\frac{\log_bx}{\log_ba}\]
anonymous
  • anonymous
you usually learn it first...
asnaseer
  • asnaseer
that is what thought @Algebraic!
asnaseer
  • asnaseer
*what I thought
anonymous
  • anonymous
hm..
anonymous
  • 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
  • anonymous
Well actually you don't have to use change of base :) .
anonymous
  • anonymous
Wait. Hold yon. You might :P .
anonymous
  • anonymous
Yeah you have too.\\
anonymous
  • 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
  • anonymous
did you get that? @Dido525
anonymous
  • anonymous
I got x =-3/2, -1 haven't checked it however...
anonymous
  • anonymous
did you get this trinomial tho: x^2+6x−9991=0
anonymous
  • anonymous
I got it!!!
anonymous
  • anonymous
No you don't get that...
anonymous
  • anonymous
I think I did it wrong
anonymous
  • anonymous
lol okay, yeah
anonymous
  • anonymous
@Algebraic! I got only -3/2 .
anonymous
  • anonymous
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3)) \] We were at that step.
asnaseer
  • asnaseer
I get 3 solutions
anonymous
  • anonymous
Use the properties of logs to expand that right log: @lucenzo
asnaseer
  • asnaseer
I think you have a typo there @Dido525 - they should all be base (x+3)
anonymous
  • anonymous
I do. I will fix.
asnaseer
  • asnaseer
\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)\]
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]
anonymous
  • anonymous
Are you there? @lucenzo
anonymous
  • anonymous
yeah. So we can expand all of the logs right?
anonymous
  • anonymous
Yep!
asnaseer
  • asnaseer
@Dido525 your last equation is wrong - you have forgotten the other log term
anonymous
  • anonymous
I am ignoring that one for now :P .
anonymous
  • anonymous
I didn't forget it.
asnaseer
  • asnaseer
then it shouldn't be set "= 4"
anonymous
  • anonymous
Yeah, my mistake.
anonymous
  • anonymous
it would be 3 then? Instead of 4
anonymous
  • 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
  • anonymous
@lucenzo
anonymous
  • anonymous
Do you notice anything special about the last log term? @lucenzo
anonymous
  • 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
  • anonymous
Good!
anonymous
  • anonymous
So you should get : \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3\]
anonymous
  • anonymous
I got to that part no problem. The next part is what confuses me xD
anonymous
  • anonymous
Now what do you notice about the first log term?
anonymous
  • 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
  • anonymous
yeah, I expanded the first term
anonymous
  • anonymous
No no!!
anonymous
  • anonymous
Remember, in a log you can bring that power down :) .
anonymous
  • anonymous
|dw:1351988960498:dw|
anonymous
  • anonymous
powering every term in the equation by the same number?
anonymous
  • anonymous
quadratic in \[\log_{x+3} (5x+9) \]
anonymous
  • anonymous
ah, that
anonymous
  • anonymous
I see now xD thanks you
anonymous
  • anonymous
u^2 -3u+2=0
anonymous
  • anonymous
So you should get : \[2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3\]
anonymous
  • anonymous
yes, I did
anonymous
  • anonymous
Now use change of base that you learned a little earlier. :) .
anonymous
  • anonymous
|dw:1351989182904:dw|
anonymous
  • anonymous
I know the normal way of using that rule where it would be: \[\log(x+3)/\log(5x+9) \]
anonymous
  • anonymous
Hint: Use it on the right log :) .
anonymous
  • anonymous
are we assuming that the "y" is just 10. so its log base of 10?
anonymous
  • anonymous
Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.
anonymous
  • anonymous
Well in this case it's base x+3 .
anonymous
  • anonymous
no, i meant for the first term. the term on the left
anonymous
  • anonymous
It's base 5x+9 .
anonymous
  • anonymous
just for the \[\log_{5x+9}(x+3) \]
anonymous
  • anonymous
The base is 5x+9 for that yes.
anonymous
  • anonymous
yeah, that's why i did \[\log(x+3)/\log(5x+9)\]
anonymous
  • anonymous
The trick for the right log is that you want to convert that right base to the base on the left log.
anonymous
  • anonymous
That's correct but you want base 5x+3 .
anonymous
  • anonymous
5x+9*
anonymous
  • anonymous
Yeah. My bad.
anonymous
  • anonymous
I got 1/\[1/\log_{5x+9}(x+3) \]
anonymous
  • anonymous
Great job!
anonymous
  • anonymous
whoops, just meant to put 1 "1/"
anonymous
  • anonymous
and now we put everything to the power of *5x+9* ?
anonymous
  • anonymous
So you should now have: \[2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)\]
anonymous
  • anonymous
Move the common terms to one side.
anonymous
  • anonymous
nvm
anonymous
  • anonymous
yeah lol, I was about to say that
anonymous
  • anonymous
It's fine :) .
asnaseer
  • asnaseer
@Dido525 - are you sure of your last step?
anonymous
  • anonymous
Yes.
asnaseer
  • asnaseer
remember you started from:\[2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9} (x+3)} =3\]
anonymous
  • anonymous
Ohh!!!!
anonymous
  • anonymous
can we not put it all to ther power of 5x+9 tho? I'm still thinking it might be simpler
anonymous
  • anonymous
oops.
asnaseer
  • asnaseer
:)
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
  • anonymous
Yep :) .
asnaseer
  • asnaseer
do you understand this step @lucenzo ?
anonymous
  • anonymous
Yeah, I'm gonna try it
anonymous
  • anonymous
I only get one solution >.> .
asnaseer
  • 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
  • anonymous
I only get 1 answer :( .
asnaseer
  • asnaseer
there are 3 answers altogether - @lucenzo have you solved the quadratic in u yet?
anonymous
  • anonymous
yeah I got u = 1 or 1/2
anonymous
  • anonymous
so 1 or 0.5
anonymous
  • anonymous
what would be the third answer?
asnaseer
  • 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
  • anonymous
right
asnaseer
  • asnaseer
so, starting with:\[log_{5x+9}(x+3)=1\]what do you get for x?
anonymous
  • anonymous
Now I got it at last :D .
asnaseer
  • asnaseer
:)
asnaseer
  • asnaseer
remember that if:\[\log_ba=c\]then:\[a=b^c\]
asnaseer
  • asnaseer
are you stuck @lucenzo ?
anonymous
  • anonymous
yeah, but I just realized I had a 2 multiplied to the log term (I'm gonna try it without the 2)
anonymous
  • anonymous
I got 3/2 for the first one (if x = 1)
asnaseer
  • 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
  • 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
  • anonymous
okay
anonymous
  • anonymous
I got -3/2 , 0 , 1.
anonymous
  • anonymous
I took the equation you had then I put both sides to the power of 5x+9 like so:
asnaseer
  • asnaseer
not quite right @Dido525
anonymous
  • anonymous
The 1 is incorrect right?
asnaseer
  • asnaseer
yes
anonymous
  • anonymous
\[5x+9^{\log_{5x+9}(x+3) } = 5x+9^1\]
anonymous
  • anonymous
erm...
asnaseer
  • asnaseer
@lucenzo you are over complicating this.
anonymous
  • anonymous
\[x+3=5x+9\]
anonymous
  • anonymous
that's right, actually.
asnaseer
  • asnaseer
yes - that is correct - but you could have got there in one step. remember:\[\log_ba=c\implies a=b^c\]
anonymous
  • anonymous
\[x-5x=9-3\]
anonymous
  • anonymous
yeah it's correct but why would you do that? :( .
anonymous
  • anonymous
That's the way we learned to write it
anonymous
  • anonymous
same thing...
anonymous
  • anonymous
Just a way of showing how to get there, it's just the same thing
asnaseer
  • asnaseer
ok - fair enough - if that is how you were taught then you should stick to that method.
anonymous
  • anonymous
I think I just showed the work
anonymous
  • anonymous
Yeah it's correct regardless.
anonymous
  • anonymous
yep you did.
asnaseer
  • asnaseer
ok so what x value do you get from this?
anonymous
  • anonymous
3/2
anonymous
  • anonymous
6/4 which reduces to 3/2
anonymous
  • anonymous
I keep getting - 3/2 hmm...
asnaseer
  • asnaseer
this step was correct:\[x-5x=9-3\]which then leads to:\[-4x=6\]
anonymous
  • anonymous
uh-oh, I made a sign error xD. I wrote -6
anonymous
  • anonymous
you're right. It's -3/2
asnaseer
  • 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
  • asnaseer
again here it would help if you listed your steps
anonymous
  • anonymous
okay
anonymous
  • anonymous
@asnaseer : I got so far: |dw:1351991548862:dw|
anonymous
  • anonymous
\[x+3=(5x+9)^{\frac{ 1 }{ 2 }}\]
anonymous
  • anonymous
|dw:1351991727301:dw|
anonymous
  • anonymous
LOL^
asnaseer
  • asnaseer
@Dido525 where did you derive that first equation from? @lucenzo that is correct - now square both sides
anonymous
  • anonymous
You get 0 and 1 if you solve for x.
anonymous
  • anonymous
If you solve x I keep getting 0 and 1.
anonymous
  • anonymous
This thread is so epic I am lagging >.< .
asnaseer
  • asnaseer
@Dido525 please let @lucenzo solve this
anonymous
  • anonymous
\[x^2 + x = 0\]
asnaseer
  • asnaseer
perfect!
asnaseer
  • asnaseer
now factorise and solve
anonymous
  • anonymous
Sorry about that anaseer.
asnaseer
  • asnaseer
np :)
anonymous
  • anonymous
I am kinda into the problem XD .
asnaseer
  • asnaseer
I know some problems are so intriguing that we feel compelled to try and solve them ourselves - it is only natural. :)
anonymous
  • anonymous
i did the quadratic formula and got x = -1 or 0
asnaseer
  • asnaseer
you have the right answer
asnaseer
  • asnaseer
but you did not need to use the quadratic formula
anonymous
  • anonymous
Yeah I got that too :) . I just had a silly algebra error.
anonymous
  • anonymous
You could have factored :P .
anonymous
  • anonymous
wha
asnaseer
  • asnaseer
\[x^2+x=0\]\[x(x+1)=0\]so\[x=0\]or:\[x=-1\]
anonymous
  • anonymous
^
anonymous
  • anonymous
-.-
asnaseer
  • asnaseer
so final solution is: x = -1.5 or -1 or 0
anonymous
  • anonymous
yeah, and I think ALL of them work. so none are extraneous
anonymous
  • anonymous
or -3/2, -1, 0 .
anonymous
  • anonymous
Yeah. I checked. All work.
asnaseer
  • asnaseer
correct - and thanks for posting such an interesting problem @lucenzo :)
anonymous
  • anonymous
I agree! I am in uni and this was the toughest log I have done in ages :) .
anonymous
  • 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
  • asnaseer
yw :) and keep up the hunger for knowledge my friend! :)
anonymous
  • anonymous
Calculus is very fun! You will enjoy it a lot :) .
anonymous
  • anonymous
Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)
anonymous
  • anonymous
Derivatives hehe...

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