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

This Question is Closed

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
\[\log_{5x+9} (x^2+6x+9)+\log_{x+3} (5x^2+24x+27) = 4\] Is that the question?
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah, that's it
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
They don't have the same base?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Darn... This complicates things...
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Okay I would factor.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3))\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Can you see anyway to simplify this? :) .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
no, I have no idea
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
You have two logs multiplied. What can you do? :) .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I meant for the right log :) .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Ohh you need help factoring? Okay: dw:1351987084250:dw
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Wait I am still in the process of solving this :P .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
and do that for both, lol okay. I'll try solving it here like that
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Wow this is the hardest log I have done so far :P .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah, same. Never seen anything like it. But I think it'll work out nicely
 2 years ago

asnaseer Group TitleBest 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.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
That was my next plan :P .
 2 years ago

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

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

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
okay, nvm then. Can you tell me what the formula is?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
But you have too O_o .
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
converting bases isn't advanced... iirc.
 2 years ago

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
you usually learn it first...
 2 years ago

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
*what I thought
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Well actually you don't have to use change of base :) .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Wait. Hold yon. You might :P .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah you have too.\\
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
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: \[x^{2} + 6x  9991 = 0\]
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
did you get that? @Dido525
 2 years ago

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I got it!!!
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
No you don't get that...
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
I think I did it wrong
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
lol okay, yeah
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
@Algebraic! I got only 3/2 .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
\[\log_{5x+9}(x+3)^{2}+\log_{x+3} ((5x+9)(x+3)) \] We were at that step.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
I get 3 solutions
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Use the properties of logs to expand that right log: @lucenzo
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I do. I will fix.
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
\[\log_{x+3} ((5x+9)(x+3))=\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Are you there? @lucenzo
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah. So we can expand all of the logs right?
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I am ignoring that one for now :P .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I didn't forget it.
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah, my mistake.
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
it would be 3 then? Instead of 4
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
So expanding all that you should get: \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9)+\log_{x+3} (x+3)=4\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Do you notice anything special about the last log term? @lucenzo
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
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"
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
So you should get : \[\log_{5x+9}(x+3)^{2}+\log_{x+3} (5x+9) =3\]
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
I got to that part no problem. The next part is what confuses me xD
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Now what do you notice about the first log term?
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah, I expanded the first term
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Remember, in a log you can bring that power down :) .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1351988960498:dw
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
powering every term in the equation by the same number?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
quadratic in \[\log_{x+3} (5x+9) \]
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
I see now xD thanks you
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
u^2 3u+2=0
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
So you should get : \[2\log_{5x+9}(x+3)+\log_{x+3} (5x+9) =3\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Now use change of base that you learned a little earlier. :) .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1351989182904:dw
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Hint: Use it on the right log :) .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
are we assuming that the "y" is just 10. so its log base of 10?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Ahh but in this case that rule is useless. Besides you can't use that unless they have the same base.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Well in this case it's base x+3 .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
no, i meant for the first term. the term on the left
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
It's base 5x+9 .
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
The base is 5x+9 for that yes.
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
The trick for the right log is that you want to convert that right base to the base on the left log.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
That's correct but you want base 5x+3 .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah. My bad.
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
whoops, just meant to put 1 "1/"
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
and now we put everything to the power of *5x+9* ?
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
So you should now have: \[2\log_{5x+9}(x+3)+1 =3\log_{5x+9} (x+3)\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Move the common terms to one side.
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah lol, I was about to say that
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
It's fine :) .
 2 years ago

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

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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
Yeah, I'm gonna try it
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I only get one solution >.> .
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
we 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.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I only get 1 answer :( .
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah I got u = 1 or 1/2
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
so 1 or 0.5
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
what would be the third answer?
 2 years ago

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Now I got it at last :D .
 2 years ago

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

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

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

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

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
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)
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I got 3/2 , 0 , 1.
 2 years ago

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
The 1 is incorrect right?
 2 years ago

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

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
\[x+3=5x+9\]
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
that's right, actually.
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
\[x5x=93\]
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
yeah it's correct but why would you do that? :( .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
That's the way we learned to write it
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
same thing...
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
Just a way of showing how to get there, it's just the same thing
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
I think I just showed the work
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah it's correct regardless.
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
yep you did.
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
6/4 which reduces to 3/2
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I keep getting  3/2 hmm...
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
uhoh, I made a sign error xD. I wrote 6
 2 years ago

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

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
@asnaseer : I got so far: dw:1351991548862:dw
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
dw:1351991727301:dw
 2 years ago

asnaseer Group TitleBest 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
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
You get 0 and 1 if you solve for x.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
If you solve x I keep getting 0 and 1.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
This thread is so epic I am lagging >.< .
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
\[x^2 + x = 0\]
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Sorry about that anaseer.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I am kinda into the problem XD .
 2 years ago

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
i did the quadratic formula and got x = 1 or 0
 2 years ago

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah I got that too :) . I just had a silly algebra error.
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
You could have factored :P .
 2 years ago

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

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

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
yeah, and I think ALL of them work. so none are extraneous
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
or 3/2, 1, 0 .
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Yeah. I checked. All work.
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
I agree! I am in uni and this was the toughest log I have done in ages :) .
 2 years ago

lucenzo Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.2
yw :) and keep up the hunger for knowledge my friend! :)
 2 years ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Calculus is very fun! You will enjoy it a lot :) .
 2 years ago

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.1
Derivatives hehe...
 2 years ago
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