Question

I have a question...

Answer 1

\[\begin{array}{l}{\LARGE \log _x}\LARGE (5{x^2})\log _5^2(x) = 1\\{\rm{Attempted:}}\\{\rm{Step 1:\;\;\;\;\;\; }}\LARGE {\log _x}(5) + {\log _x}{(x)^2} \cdot {\log _5}(x) \cdot {\log _5}(x) = 1\\{\rm{Step 2: \quad }}\LARGE {\log _x}(5) + 2 \cdot {\log _x}(x) \cdot {\log _5}(x) \cdot {\log _5}(x) = 1\\{\rm{Step 3: \quad }}\LARGE {\log _x}(5) + 2 \cdot 1 \cdot {\log _5}(x) \cdot {\log _5}(x) = 1\\\end{array}\]
or maybe we should just let it:
\[\Large \log _5^2(x)\]
I'm pretty sure I've broken some rules till here...

Multiple choices are:

\[\LARGE a) \quad \quad x=\sqrt5 \quad ,\quad x=\frac15\]\[\LARGE b) \quad x=\frac12 \quad ,\quad x=\sqrt2 \]\[\LARGE c) \quad x=-\frac12 \quad ,\quad x=\sqrt2 \]\[\LARGE d) \quad x=-\sqrt5 \quad ,\quad x=-\frac15\]
as much as I see, ti's going be a a quadratic equation...
but how do I continue ?

Answer 2

O_O Kershnik have a question..

Answer 3

ehhehehehehe lol

Answer 4

:)

Answer 5

what was that?? solve for x??

Answer 6

well I guess it is ... in my book it only says :
The solutions of this logarithm equation are... and there are multiple choices :(

Answer 7

substitute one by one each choice and see at which one your equation is satisfied.

Answer 8

eeehh.. nice one @Shayaan_Mustafa But what if in the future there are no multiple choices given ? O_O lol ;) thanks for the advice :P

Answer 9

then you have to solve it. hehehe.. lolololololololololoooolololololol....

Answer 10

hahahaa. ;)

Answer 11

thats it ... you don't have to expand it ... and make it ugly. And your attempted step 1 is wrong

should have been:
\( \ (log_x(5)+log_x(x^2))⋅log_5(x)⋅log_5(x)=1 \)

Answer 12

huh.. ? ok, Let me try, let's see how far can I go, (I'll be stuck on the second step I'm sure) LOL haha...

Answer 13

Here's the photo if you would like to see it :)

Answer 14

I told you I'll be stuck on the second step :( ...
Even if we leave it in bracket like this:
\[\LARGE [\log_x(5)+2]\cdot \log _5x \cdot \log_5 x-1=0\]
we still have logs with different bases how do I multiply now ?

Answer 15

multiply them

Answer 16

I don't know how :$

Answer 17

@experimentX do you mind if I call anyone? :$

Answer 18

I don't mind ...
OS crashed

Answer 19

auff... I know your pain. Mine one crashes too :F

Answer 20

\(\huge \log_5x = \frac{\log_ax}{\log_a5}\) you need to use this property.

Answer 21

I will make a little simpler.

Answer 22

\[\LARGE \log_ab=\frac{1}{\log_ba}\] is this a rule, or my brain is getting old ! ?

Answer 23

experimentX is right.

Answer 24

It is i guess ... wow I didn't know that.

Answer 25

does it help in this case.. ? I thought about checking algebra cheat sheet (I always do), but there's not much about logarithms :S... http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCkQFjAA&url=http%3A%2F%2Ftutorial.math.lamar.edu%2Fpdf%2FAlgebra_Cheat_Sheet.pdf&ei=JfSFT8-cF8XAtAb6pOC7Bg&usg=AFQjCNGNmumjdKSokx8p6h5l0hy4UVinmA&sig2=bGGZ6kPYD-WDeUsFEmEWVw

Answer 26

experiment... If I use your rule, which number will be instead of "a" ? :( I still don't get how should I apply that rule :(

Answer 27

@JamesJ @amistre64 @dumbcow @Diyadiya @radar I need help please...

Answer 28

It doesn't matter which no you choose,

Answer 29

http://upload.wikimedia.org/wikipedia/en/math/0/3/a/03a7c4f8461ee26493b9cf547a82e390.png

Answer 30

get everyone speaking in the same log for starters is my first thought

Answer 31

after you do that ... you will have quadratic equation in \( \large \log_5x\) ... i guess it would be easy then.

Answer 32

I think I just found a solution... Let me type it :F..

Answer 33

\[\log _x (5{x^2})\log _5^2(x) = 1 \]
\[\log _x (5{x^2}) = \frac{1}{\log _5^2(x)} \]
\[\frac{ln(5{x^2})}{ln(x)} = \frac{1}{\frac{ln^2(x)}{ln^2(5)}} \]
\[\frac{ln(5{x^2})}{ln(x)} = \frac{ln^2(5)}{ln^2(x)} \]
\[ln(5{x^2}) = \frac{ln^2(5)}{ln(x)} \]
so far is what ive got but this thing is terrible on IE

Answer 34

\[\LARGE \log_x(5x^2)\cdot \log_5^2x=1\]
\[\LARGE (\log_x5+\log _xx^2 )\cdot \log_5^2x=1\]
\[\LARGE (\log_x5+2 )\cdot \log_5^2x=1\]
\[\LARGE \left(\frac{1}{\log_5x }+2 \right)\cdot \log_5^2x=1\]
\[\LARGE \left(\frac{1}{\log_5x }+\frac{2\cdot \log_5x}{\log_5x } \right)\cdot \log_5x \cdot \log_5x=1\]
\[\LARGE \left(\frac{2 \cdot \log_5x+1 }{\log_5x } \right)\cdot \log_5x \cdot \log_5x=1\]
\[\LARGE \left(2 \cdot \log_5x+1 \right)\cdot \log_5x=1\]

so far is it good?

Answer 35

let \( \huge \log_5x = y\)

Answer 36

ok... let me see what I can do :)

Answer 37

\[2ln(5x) \to \ 2ln(5)+2ln(x)\]

\[ln(x)(2ln(5)+2ln(x)) = ln^2(5)\]

\[2ln(5)ln(x)+2ln^2(x) = ln^2(5)\]

\[2ln^2(x)+2ln(5)ln(x)- ln^2(5)=0\]

ln(x) = r; and say ln^2(5) is some arbitrary constant to clean it up a little

\[2r^2+2r- K=0\]

Answer 38

lost the ln(5) in the middle during typing :)

Answer 39

@amistre64 @experimentX I do really really appreciate your help ... I'm typing my version :F lol let's see how many rules I've broken so far :P lol

Answer 40

\[\LARGE (2y+1)\cdot y=1\]
\[\LARGE 2y^2+y-1=0\]
\[\LARGE x_{1/2}=\frac{-1\pm\sqrt{1+8}}{4}\]

\[\LARGE x_{1}=\frac{-1-3}{4}=-1\]


\[\LARGE x_{1/2}=\frac{-1+3}{4}=\frac12\]

now let's substitute back:

\[\LARGE \log_5x=-1\]
\[\LARGE x_1=\frac15 \]

\[\LARGE \log_5x=\frac12\]
\[\LARGE x_2=\sqrt5\]

I think option A must be... (If I'm not wrong :( )

Answer 41

it is ...

Answer 42

huh... I feel like I just took off a sack of sand off my back LOL.... thank you everyone for the help, I hope I can help you out one day, even if that day never comes, I won't forget these ones ;) thank you

Answer 43

I'm closing it. Bye bye :)

Answer 44

sure np