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anonymous
 3 years ago
I need to solve for x: log(base 5x + 9) (x^2 + 6x + 9) + log(base x+3) (5x^2 + 24x + 27) = 4
anonymous
 3 years ago
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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4\] Is that the question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0They don't have the same base?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Darn... This complicates things...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you see anyway to simplify this? :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You have two logs multiplied. What can you do? :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This 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
 3 years ago
Best ResponseYou've already chosen the best response.0They don't have the same base so you can't do "loga + logb = log(a*b)"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I meant for the right log :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh you need help factoring? Okay: dw:1351987084250:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You do log (5x^2 + 24x + 27)/(x+3)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait I am still in the process of solving this :P .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and do that for both, lol okay. I'll try solving it here like that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wow this is the hardest log I have done so far :P .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, same. Never seen anything like it. But I think it'll work out nicely

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@Dido525 your approach is correct. think of making use of the change of base for logs you'll be there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That was my next plan :P .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@lucenzo do you know about the change of base formula for logs?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no. I think I might have to solve this question w/o using the formula, since we never learned it

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I can't see how to solve this without using that formula?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay, nvm then. Can you tell me what the formula is?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But you have too O_o .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I 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
 3 years ago
Best ResponseYou've already chosen the best response.0converting bases isn't advanced... iirc.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2change of base is given by:\[\log_ax=\frac{\log_bx}{\log_ba}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you usually learn it first...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2that is what thought @Algebraic!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, I've used that formula LOL. I actually was thinking that from the beginning. In this case I did \[\log_{a}n \] first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well actually you don't have to use change of base :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait. Hold yon. You might :P .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then 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
 3 years ago
Best ResponseYou've already chosen the best response.0did you get that? @Dido525

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got x =3/2, 1 haven't checked it however...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did you get this trinomial tho: x^2+6x−9991=0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No you don't get that...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I did it wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Algebraic! I got only 3/2 .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3)) \] We were at that step.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use the properties of logs to expand that right log: @lucenzo

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I think you have a typo there @Dido525  they should all be base (x+3)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Are you there? @lucenzo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah. So we can expand all of the logs right?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@Dido525 your last equation is wrong  you have forgotten the other log term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am ignoring that one for now :P .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2then it shouldn't be set "= 4"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it would be 3 then? Instead of 4

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So expanding all that you should get: \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you notice anything special about the last log term? @lucenzo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, 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
 3 years ago
Best ResponseYou've already chosen the best response.0So you should get : \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got to that part no problem. The next part is what confuses me xD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now what do you notice about the first log term?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\log_{5x+9} (x+3)^2 = \frac{ 2}{\log_{x+3} (5x+9) }\] \[\log_{x+3} (5x+9)(x+3) = 1 +\log_{x+3} (5x+9)\] \[\log_{x+3} (5x+9)+\frac{ 2}{\log_{x+3} (5x+9) }=3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, I expanded the first term

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Remember, in a log you can bring that power down :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351988960498:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0powering every term in the equation by the same number?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0quadratic in \[\log_{x+3} (5x+9) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I see now xD thanks you

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you should get : \[2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now use change of base that you learned a little earlier. :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351989182904:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I know the normal way of using that rule where it would be: \[\log(x+3)/\log(5x+9) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: Use it on the right log :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are we assuming that the "y" is just 10. so its log base of 10?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well in this case it's base x+3 .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no, i meant for the first term. the term on the left

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just for the \[\log_{5x+9}(x+3) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The base is 5x+9 for that yes.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, that's why i did \[\log(x+3)/\log(5x+9)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The trick for the right log is that you want to convert that right base to the base on the left log.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's correct but you want base 5x+3 .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got 1/\[1/\log_{5x+9}(x+3) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whoops, just meant to put 1 "1/"

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and now we put everything to the power of *5x+9* ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you should now have: \[2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Move the common terms to one side.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah lol, I was about to say that

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@Dido525  are you sure of your last step?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2remember you started from:\[2\log_{5x+9}(x+3)+\frac{1}{\log_{5x+9} (x+3)} =3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can we not put it all to ther power of 5x+9 tho? I'm still thinking it might be simpler

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2from 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2do you understand this step @lucenzo ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, I'm gonna try it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I only get one solution >.> .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2we got to:\[2u+\frac{1}{u}=3\]multiply both sides by u to get:\[2u^2+1=3u\]\[2u^23u+1=0\]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
 3 years ago
Best ResponseYou've already chosen the best response.0I only get 1 answer :( .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2there are 3 answers altogether  @lucenzo have you solved the quadratic in u yet?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I got u = 1 or 1/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what would be the third answer?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2that is correct. so now you need to solve:\[log_{5x+9}(x+3)=1\]and:\[log_{5x+9}(x+3)=0.5\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2so, starting with:\[log_{5x+9}(x+3)=1\]what do you get for x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now I got it at last :D .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2remember that if:\[\log_ba=c\]then:\[a=b^c\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2are you stuck @lucenzo ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, but I just realized I had a 2 multiplied to the log term (I'm gonna try it without the 2)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got 3/2 for the first one (if x = 1)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2plz 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
 3 years ago
Best ResponseYou've already chosen the best response.2it might be better if you showed your steps here so that we can help you spot where you may have made a mistake

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I took the equation you had then I put both sides to the power of 5x+9 like so:

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2not quite right @Dido525

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 1 is incorrect right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[5x+9^{\log_{5x+9}(x+3) } = 5x+9^1\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@lucenzo you are over complicating this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's right, actually.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2yes  that is correct  but you could have got there in one step. remember:\[\log_ba=c\implies a=b^c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah it's correct but why would you do that? :( .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's the way we learned to write it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Just a way of showing how to get there, it's just the same thing

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2ok  fair enough  if that is how you were taught then you should stick to that method.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I just showed the work

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah it's correct regardless.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2ok so what x value do you get from this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06/4 which reduces to 3/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I keep getting  3/2 hmm...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2this step was correct:\[x5x=93\]which then leads to:\[4x=6\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhoh, I made a sign error xD. I wrote 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're right. It's 3/2

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2ok, so first solution is x=1.5. next, you need to solve this for x:\[log_{5x+9}(x+3)=0.5\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2again here it would help if you listed your steps

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@asnaseer : I got so far: dw:1351991548862:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x+3=(5x+9)^{\frac{ 1 }{ 2 }}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351991727301:dw

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@Dido525 where did you derive that first equation from? @lucenzo that is correct  now square both sides

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You get 0 and 1 if you solve for x.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you solve x I keep getting 0 and 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This thread is so epic I am lagging >.< .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2@Dido525 please let @lucenzo solve this

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2now factorise and solve

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry about that anaseer.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am kinda into the problem XD .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2I know some problems are so intriguing that we feel compelled to try and solve them ourselves  it is only natural. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i did the quadratic formula and got x = 1 or 0

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2you have the right answer

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2but you did not need to use the quadratic formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I got that too :) . I just had a silly algebra error.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You could have factored :P .

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2\[x^2+x=0\]\[x(x+1)=0\]so\[x=0\]or:\[x=1\]

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2so final solution is: x = 1.5 or 1 or 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, and I think ALL of them work. so none are extraneous

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah. I checked. All work.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.2correct  and thanks for posting such an interesting problem @lucenzo :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I agree! I am in uni and this was the toughest log I have done in ages :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol. 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
 3 years ago
Best ResponseYou've already chosen the best response.2yw :) and keep up the hunger for knowledge my friend! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Calculus is very fun! You will enjoy it a lot :) .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, I heard that it has less writing and that you learn 'tricks' for solving questions ALOT quicker (like never before)
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