anonymous
  • anonymous
\[\Large \log^2x^3-20\log\sqrt x +1=? \] How many solutions does equation have? 1 2 3 4
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I can't start it :(
anonymous
  • anonymous
I know that \[\Large \log^2x =2\log x\] and \[\Large \log_{x^t}a=\frac 1t \log_xa\] and something else but as much as I see I don't need the second one here ... :S Any hint??
Mertsj
  • Mertsj
What is it equal to? If it is an equation, it must be equal to something.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
sorry... equal to 0 :$ my bad
anonymous
  • anonymous
I thought that I write that :$
Mertsj
  • Mertsj
\[\log_{10}x^2=2\log_{10}x \]
Mertsj
  • Mertsj
\[(\log_{10}x )^{2}\neq2\log_{10}x \]
anonymous
  • anonymous
\[\Large \log^2x =\log x\cdot \log x ??\]
Mertsj
  • Mertsj
Yes
TuringTest
  • TuringTest
\[\Large \log^2x^3-20\log\sqrt x +1=9\log^2x-10\log x+1\]it's not equal to 0 is it?cause then it would be quadratic in log x
TuringTest
  • TuringTest
\[\Large \log^2x^3-20\log\sqrt x +1=0\]would make this problem a little more direct
anonymous
  • anonymous
Is given: \[\Large \log^2x^3-20\log\sqrt x+1=0\]
TuringTest
  • TuringTest
sweet, then you can use the quadratic formula from where I got you too
TuringTest
  • TuringTest
\[\log^2x^3-20\log\sqrt x +1=9\log^2x-10\log x+1=0\]substitute\[\log x=y\]
anonymous
  • anonymous
what about \[\Large \log \sqrt x\]
anonymous
  • anonymous
Ufff.. I see sorry Let me try.. I'll see what I can do ;)
Mertsj
  • Mertsj
Turing, sorry to be so slow, but how is log^2x^3=9log^2x
TuringTest
  • TuringTest
not at all :)\[(\log x^3)^2=(3\log x)^2=9\log x\]
Mertsj
  • Mertsj
Oh. I got it. Sorry.
TuringTest
  • TuringTest
forgot the log^2 in the last part...
Mertsj
  • Mertsj
Now it factors!!!
TuringTest
  • TuringTest
sweet jeezus we might be able to solve it :D
anonymous
  • anonymous
So I have... \[\Large x_{1/2}\frac{10\pm\sqrt{100-36}}{18}\] And \[\Large x_{1/2}\frac{10\pm 8}{18}\longrightarrow x_1=1 \quad ,x_2=\frac19\] Right?
TuringTest
  • TuringTest
I don't know, I don't feel like checking mertsj says it factors. let's see what he gets
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
not factorable? ok let me see what I get...
Mertsj
  • Mertsj
\[(9logx-1)(logx-1)=0\]
TuringTest
  • TuringTest
I though it factored :) so it looks like we have the right answer... (for the log part at least)
TuringTest
  • TuringTest
I suppose to finish this out we need to know what base the log is is it 10?
anonymous
  • anonymous
Yes .. because there's no base given ...
Mertsj
  • Mertsj
So 10^(1/9) and 10
TuringTest
  • TuringTest
\[\log x=\frac19\implies x=\sqrt[9]10\]ew... ugly number well, ok not that ugly, but still
TuringTest
  • TuringTest
at least the 10 is pretty :)
anonymous
  • anonymous
LOL ... thanks guys so much, thanks in advance
Mertsj
  • Mertsj
Very much so. Do these answers have to be checked in the original?
Mertsj
  • Mertsj
Thank Turing. He's the brains of the outfit.
TuringTest
  • TuringTest
Welcome, thanks for the compliment @Kreshnik why do you say "in advance" ? you assume I am going to keep helping that way? lol
TuringTest
  • TuringTest
fyi you usually say "thanks in advance" before the person has actually done the favor
TuringTest
  • TuringTest
(English lesson, lol)
Mertsj
  • Mertsj
Oh. It only wanted to know how many solutions there are.
Mertsj
  • Mertsj
A true liberal arts scholar!!
TuringTest
  • TuringTest
haha^ screw that "how may" business, once you recognize it's quadratic in logx you know there will be 2 solutions, so I guess we could have stopped there
TuringTest
  • TuringTest
...so long as the discriminant is not zero
anonymous
  • anonymous
LOL ... hahaha , Since my own language it's not English , I've seen in a forum a guy who said that: Thanks in advance .... uhauhuha LOL INDEED he said that before the answer it was given, sorry if I made any mistake , I'm trying my best, I'm 18 years old, and I'm from Kosovo but I know English too (at least I think I do ! LOL ) hahaha.. @TuringTest thanks for English lessons :) ... can't stop laughing !
TuringTest
  • TuringTest
That's cool, I'm learning Spanish and I teach English, so I know your pain
Mertsj
  • Mertsj
Or one of the answers is not negative.
TuringTest
  • TuringTest
good point @Mertsj
Mertsj
  • Mertsj
You teach English??? And you're such a math whiz. Do you teach math too?
anonymous
  • anonymous
I know spanish too... (I understand pretty much !) but I know English better !! ... (I don't know to write Spanish at all !)
TuringTest
  • TuringTest
I am a tutor in English and math for the school where I am studying Spanish here in Mexico I actually suck at grammatical rules and such, but I get by being a native speaker.
Mertsj
  • Mertsj
Be safe. And don't get involved in the drug wars!!
TuringTest
  • TuringTest
No worries, that's a border problem. I'm in Guadalajara, to the south-central. Thanks though, and take care yourself :)
Mertsj
  • Mertsj
Thanks. Will do.

Looking for something else?

Not the answer you are looking for? Search for more explanations.