Question

\[\Large \log^2x^3-20\log\sqrt x +1=? \]

How many solutions does equation have?

1
2
3
4

Answer 1

I can't start it :(

Answer 2

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??

Answer 3

What is it equal to? If it is an equation, it must be equal to something.

Answer 4

sorry... equal to 0 :$ my bad

Answer 5

I thought that I write that :$

Answer 6

\[\log_{10}x^2=2\log_{10}x \]

Answer 7

\[(\log_{10}x )^{2}\neq2\log_{10}x \]

Answer 8

\[\Large \log^2x =\log x\cdot \log x ??\]

Answer 9

Yes

Answer 10

\[\Large \log^2x^3-20\log\sqrt x +1=9\log^2x-10\log x+1\]it's not equal to 0 is it?cause then it would be quadratic in log x

Answer 11

\[\Large \log^2x^3-20\log\sqrt x +1=0\]would make this problem a little more direct

Answer 12

Is given:

\[\Large \log^2x^3-20\log\sqrt x+1=0\]

Answer 13

sweet, then you can use the quadratic formula from where I got you too

Answer 14

\[\log^2x^3-20\log\sqrt x +1=9\log^2x-10\log x+1=0\]substitute\[\log x=y\]

Answer 15

what about \[\Large \log \sqrt x\]

Answer 16

Ufff.. I see sorry
Let me try.. I'll see what I can do ;)

Answer 17

Turing, sorry to be so slow, but how is log^2x^3=9log^2x

Answer 18

not at all :)\[(\log x^3)^2=(3\log x)^2=9\log x\]

Answer 19

Oh. I got it. Sorry.

Answer 20

forgot the log^2 in the last part...

Answer 21

Now it factors!!!

Answer 22

sweet jeezus we might be able to solve it :D

Answer 23

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?

Answer 24

I don't know, I don't feel like checking
mertsj says it factors. let's see what he gets

Answer 25

ok

Answer 26

not factorable?
ok let me see what I get...

Answer 27

\[(9logx-1)(logx-1)=0\]

Answer 28

I though it factored :)
so it looks like we have the right answer... (for the log part at least)

Answer 29

I suppose to finish this out we need to know what base the log is
is it 10?

Answer 30

Yes .. because there's no base given ...

Answer 31

So 10^(1/9) and 10

Answer 32

\[\log x=\frac19\implies x=\sqrt[9]10\]ew... ugly number
well, ok not that ugly, but still

Answer 33

at least the 10 is pretty :)

Answer 34

LOL ... thanks guys so much, thanks in advance

Answer 35

Very much so. Do these answers have to be checked in the original?

Answer 36

Thank Turing. He's the brains of the outfit.

Answer 37

Welcome, thanks for the compliment
@Kreshnik why do you say "in advance" ?
you assume I am going to keep helping that way? lol

Answer 38

fyi you usually say "thanks in advance" before the person has actually done the favor

Answer 39

(English lesson, lol)

Answer 40

Oh. It only wanted to know how many solutions there are.

Answer 41

A true liberal arts scholar!!

Answer 42

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

Answer 43

...so long as the discriminant is not zero

Answer 44

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 !

Answer 45

That's cool, I'm learning Spanish and I teach English, so I know your pain

Answer 46

Or one of the answers is not negative.

Answer 47

good point @Mertsj

Answer 48

You teach English??? And you're such a math whiz. Do you teach math too?

Answer 49

I know spanish too... (I understand pretty much !) but I know English better !! ... (I don't know to write Spanish at all !)

Answer 50

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.

Answer 51

Be safe. And don't get involved in the drug wars!!

Answer 52

No worries, that's a border problem.
I'm in Guadalajara, to the south-central.
Thanks though, and take care yourself :)

Answer 53

Thanks. Will do.