Question

Given:
\[\Huge x^{[-4+\log_3(x)]}={1\over 27}\]
Attempted:
\[\Huge x^{[-4+\log_3(x)]}=3^{-3 }\]
\[\huge [-4+\log_3(x)]\cdot \log (x)=-3\cdot \log(3)\]
Pretty dumb attempt huh?

Multiple choices are:


A. \[\LARGE x_1=2.4 \quad ,\quad x_2=2\]

B. x=3 ,x=27


C.
x=-8 , x=2


D.
\[\LARGE x=\frac19 \quad ,\quad x=3\]
any hint would be great !

Answer 1

Is it B?

Answer 2

LOL, I'm not trying to see who's smarter. I don't know the answer :$. I was hoping someone to help me do it :(

Answer 3

Oh sorry, I thought you had the answer... Forgive me..
well I did it like this
\[x^{-4+log_{3}x} = \frac{1}{27}\]\[\frac{x^{log_{3}x}}{x^4} = \frac{1}{27}\]\[x^{log_{3}x} = \frac{x^4}{27}\]
Then, I did it by observation...

Answer 4

... give me a minute, I'll try something, although I'm sure I'll screw it up :F

Answer 5

starting from your second step:
\[\Huge {x^{\log _{3}x}\over x^4}=\frac{1}{27}\]
\[\Huge 3^3\cdot x^{\log _{3}x}=x^4 \]

Now I was thinking about:
\[\Huge \log_ab=\frac{1}{\log_ba}\]
and
\[\Huge x^{\log_xa}=a\]
to change exponent of X
like...
\[\Huge x^{\log_3x}=x^{1 \over \log_x3}\]
But I don't know how it would be, since there's a fraction. ... :(
maybe :
\[\Huge x^{(\log_x3)^{-1}}\]something...

Answer 6

If I could only get rid of that log exponent that would be much easier ...

Answer 7

Hmm.. You're now doing something I haven't learnt...

Answer 8

that makes no sense anyway...
let's assume we have:
\[\LARGE 3^3\cdot 3^{-1}=x^4\]
\[\LARGE 3^2=x^4\]
\[\LARGE x=\pm \sqrt3 \]

Answer 9

LOL , I'm trying to do something I haven't learnt too, because I don't see other way :B hahah...

Answer 10

anyway Log rules still hold ... I didn't invent them !

Answer 11

I'm lost :S
Perhaps you can sub the values into the equation to get the answer (This is always the ultimate way for MC question :| )

Answer 12

I want to know how to do it, anyway... Time to call some1 :F .. thanks a lot ;) @Callisto

Answer 13

Sorry that I can't help :(

Answer 14

ahha.. If you feel bad that you can't help me, I hate to think how should I feel when I know I never helped you :F

Answer 15

@satellite73 @Ishaan94 ... please help me ...

Answer 16

wolframa says that answer is B.. :(

Answer 17

here is my try
first step i guess is take the log (base 3) as you did and write
\[(-4+\log_3(x))\log_3(x)=-3\]
then multiply out and get
\[(\log_3(x))^2-4\log_3(x)+3=0\] solve the quadratic get
\[(\log_3(x)-1)(\log_3(x)-3)=0\]
\[\log_3(x)=1\implies x = 3\]
\[\log_3(x)=3\implies x=27\]

Answer 18

i didn't check the answers but it looks good
if this is one of the choices go with that

Answer 19

*Big applause to satellite73*

Answer 20

(blush)

Answer 21

clap clap clap... :F

Answer 22

I'll just write it down ..

Answer 23

whoaaa... Sooo sick. I knew you'd help me. Thank you everyone. I'm closing it :)

Answer 24

I'm sure I'm a dumb after reading satellite73's solution :(
Thanks for posting this question!

Answer 25

.. me too, never thought about taking logs in base 3 :F ... can't stop laughing LOL. @Callisto thank you for your time. and everyone who at least thought about helping me. ;)

Answer 26

my solution is....