Question

express y as a function of x. what is the domain?
a) log 3 + log y = log (x+2) - log x

this is what i did:
log 3 + log y = log (x+2) - log x
log y = log (x+2) - log x - log 3
log y = log (x+2) - log (3x)
log y = log [(x+2)/(3x)]
y = (x+2)/(3x)

.. is this right?
or did i do something wrong..
my friend did it another way.. so idk

Answer 1

yes you're right

Answer 2

just to clarify.. if i have log on both the right and left side... do i just "remove it" ??

Answer 3

yep

Answer 4

however if you have something like
\[\log y = x\]
it will become
\[y = 10^x\]

Answer 5

okay.
so i'm given another equation where log 4y = x + log 4...
would this become log 4y = 10 x + log 4 ?
then log 4y = log 40x ... ? :S

Answer 6

\[\huge x \implies x(1) \implies x(\log 10)\]

Answer 7

SO... it would be log 4y = x(log10) + log 4

Answer 8

?*

Answer 9

yep

Answer 10

you can combine it further

Answer 11

log 4y = x(log40) ?

Answer 12

no..

Answer 13

\[x \log 10 \implies \log 10^x\]

Answer 14

uhm.. \[\log 4y = \log 10^{x} + \log 4\]\[\log 4y = \log (10^{x} + 4)\]

Answer 15

10^x + 4?

Answer 16

40^x ?

Answer 17

my bad.. i meant 40^x

Answer 18

still no

Answer 19

\[\huge a^m \times b \ne (ab)^m\]

Answer 20

10^x * 4 is 10^x *4....you cant do anything about it

Answer 21

... oh my gosh.. sorry! okay so
it would then be log 4y = log(10^x(4))
4y = (10^x(4))
y = 10^x

Answer 22

yes!!

Answer 23

i am so sorry... ! im just making this a lot harder than it should be... thank you so much !