Question

log[2](x)((1)/(9))=-2 answer plz the viarable is x

Answer 1

Could you write this out to look more simple? :P

Answer 2

\(\large \log_{2} \dfrac{1}{9}=-2\)
where is the variable ?? O.o

Answer 3

Is there a variable missing: log[2]((1)/(9))=-2 answer plz

Answer 4

$$
\huge log_2\pmatrix{\cfrac{1}{9}}= -2
$$

Answer 5

That yields 1/4 = 1/9, doesn't it?

Answer 6

They don't equal,hence the missing variable.

Answer 7

@sidney38

Answer 8

I think this is like a rearrange kinda thing... So no variables in this log.

Answer 9

There needs to be because 1/4= 1/9 is a false statement.

Answer 10

The only way to make a true statement would be to solve for x.

Answer 11

variable is x yeah so varible is missing

Answer 12

@sidney can you take a screenshot of it?

Answer 13

^ you tagged different sidney :P

Answer 14

hahaha.

Answer 15

@sidney38

Answer 16

ohh shoot, wrong sidney darn

Answer 17

she messaged me that the variable is x
maybe she is unable to post here....

Answer 18

refresh page...

Answer 19

it works if it a base 3 log.... otherwise its false.

Answer 20

\[\log_{3} \frac{1}{9} = \log_{3} 3^{-2} = -2\]

Answer 21

True enough.

Answer 22

now i can see a 'x'

Answer 23

yes, the "x" would be the \(3^{-2}\)

Answer 24

\(\large \log_2{\frac{x}{9}}=-2 \)

Answer 25

\[\large 2^{-2} = \cfrac{x}{9}\]\[\large \cfrac{1}{4}=\cfrac{x}{9} \]

Answer 26

\(\huge \log_ab=c \implies b=a^c\)

Answer 27

\(\large \log_2{\frac{x}{9}}=-2\)
\(\implies\) what jhannybean posted...

Answer 28

$$
log_2\pmatrix{\cfrac{x}{9}}= -2\\
2^{log_2\pmatrix{\cfrac{x}{9}}} = 2^{-2}\\
\text{using the log cancellation rule}\\
\color{blue}{a^{log_ax} = x}\\
2^{-2} = \cfrac{x}{9}\\
\cfrac{1}{4}=\cfrac{x}{9}
$$
as shown above by Jhannybean