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what's the original question ?

Log\/2(3)+log\/2(x)=3

i mean equation \[\log_2(3)+\log_2 x=3\] like this OR

\[\frac{ \log_2 3 }{ \log_2 x} =3\]

maybe or maybe division hmm
which one is correct ? graysongraddyLOL

the first one is the original

oh okay familiar with the log properties ??

Not exactly.

look at that and tell me which one you should apply ?

The last one.?

Math just absolutely dumbfounds me.

hmm alright there is PLUS sign between log
now look at the log identities

the subexponent of 2 is the identity correct.?

2 is base

so with the equation that i am using, How do i get a Log on the other side of the equal sign.?

So multiply 3 and x.?

\[\huge\rm log_2(3)+\log_2 x=3\] like this OR
yes right!!

|dw:1439575720373:dw|

That just absolutely confused me.

well you will see how easy it's
just have to move that variables around

so log2 (3) = 3x.?

we need to convert to exponential there shouldn't be an log
b=2
how would write 2 with 3 exponent ?

2 to the 3 ppower

so 2^3=3?

2^3 = 3x not just x

now solve for x

\[x=\frac{ 8 }{ 3 }\]

:)

Okay cool :)

Can you give me a example logarithm so i can see if i got the hang of it.?

why not!!

Thank you so much :D

\[\huge\rm log_3 (4) - \log_3 (y) = 2\] look at the log properties

So id use the first one right.?

yea quotient property

then it would go to \[\log_{3}(4 \div x)=3\]

yep right

it's y not x but it's okay :P

i'm not gonna take off points like teachers do lol

Oh whoops lol

So it would go to \[3^{4}=3x\]

hmm no

Where did i go wrong.?

try again!

okay so it would go to \[\frac{ 4 }{ x } = 2x \]

how did you get two x's ?

wait it would just be 2 on the right side

then i multiply 4 from the left side to get 8 on the right side so x=8.?

|dw:1439577454761:dw|

\[3^{2} =4x.?\]

not really hmm
it's 4/x so that would stay same at right side \[3^2=\frac{ 4 }{ x }\]now solve for x

\[x=\frac{ 4 }{ 9 }\]

yep!

Thank you so much!

my pleasure!