Check this.? Im not very good with logarithms and i watched a video and it said to do it like this but im not sure
Log\/2(3)+log\/2(x)=3
then cancel out the logs as they are the same base so,
3+x=3
-3 -3
x=0???

- anonymous

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- Nnesha

what's the original question ?

- anonymous

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

- Nnesha

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

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## More answers

- Nnesha

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

- Nnesha

maybe or maybe division hmm
which one is correct ? graysongraddyLOL

- anonymous

the first one is the original

- Nnesha

oh okay familiar with the log properties ??

- anonymous

Not exactly.

- Nnesha

quotient rule\[\large\rm log_b x - \log_b y = \log_b \frac{ x }{ y}\]
to condense you can change subtraction to division
product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\]
addition ----> multiplication
power rule \[\large\rm log_b x^y = y \log_b x\]

- Nnesha

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

- anonymous

The last one.?

- anonymous

Math just absolutely dumbfounds me.

- Nnesha

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

- anonymous

the subexponent of 2 is the identity correct.?

- Nnesha

2 is base

- Nnesha

quotient rule\[\large\rm\color{reD}{ log_b x - \log_b y = \log_b \frac{ x }{ y}}\]
to condense you can change subtraction to division
product rule \[\large\rm \color{reD}{log_b x + \log_b y = \log_b( x \times y )}\]
addition ----> multiplication
power rule \[\large\rm \color{Red}{log_b x^y = y \log_b x}\]
you don't need ^power rule for this e quation
we should apply it when there is a number at front of log

- anonymous

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

- Nnesha

alright we need to apply product rule which is \[\large\rm \color{reD}{log_b x + \log_b y = \log_b( x \times y )}\]
log_b same base so take it out and multiply x y
\[\log_b x+ \log_b y\]
\[\log_b (x \times y)\]

- anonymous

So multiply 3 and x.?

- Nnesha

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

- Nnesha

that's how we should condense log equations
\[\huge\rm log_2 (3 \times x)=3\]
now we should convert log to exponential form

- Nnesha

|dw:1439575720373:dw|

- anonymous

That just absolutely confused me.

- Nnesha

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

- Nnesha

right side term becomes exponent of base
and move the left side term to the right side |dw:1439575981638:dw|

- anonymous

so log2 (3) = 3x.?

- Nnesha

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

- Nnesha

2 to the 3 ppower

- anonymous

so 2^3=3?

- Nnesha

2^3 = 3x not just x

- Nnesha

now solve for x

- anonymous

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

- Nnesha

:)

- anonymous

Okay cool :)

- anonymous

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

- Nnesha

why not!!

- anonymous

Thank you so much :D

- Nnesha

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

- anonymous

So id use the first one right.?

- Nnesha

yea quotient property

- anonymous

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

- Nnesha

yep right

- Nnesha

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

- Nnesha

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

- anonymous

Oh whoops lol

- anonymous

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

- Nnesha

hmm no

- anonymous

Where did i go wrong.?

- Nnesha

the number which is opposite of the log supposed to be the exponent of base \[\log_3 ( \frac{ 4 }{ x }) =2\]
and btw it's equal to 2
i would take off points for this mistake

- Nnesha

try again!

- anonymous

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

- Nnesha

how did you get two x's ?

- anonymous

wait it would just be 2 on the right side

- anonymous

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

- Nnesha

|dw:1439577454761:dw|

- Nnesha

here is the example to convert log to exponential form \[\huge\rm log_\color{ReD}{b} x = \color{blue}{y}\] b is base \[\huge\rm \color{ReD}{b}^\color{blue}{y} =x\]

- anonymous

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

- Nnesha

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

- anonymous

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

- Nnesha

yep!

- anonymous

Thank you so much!

- Nnesha

my pleasure!

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