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???

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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???

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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

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\[\frac{ \log_2 3 }{ \log_2 x} =3\]
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.
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 \[\large\rm log_b x^y = y \log_b x\]
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
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
so with the equation that i am using, How do i get a Log on the other side of the equal sign.?
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 \[\log_b x+ \log_b y\] \[\log_b (x \times y)\]
So multiply 3 and x.?
\[\huge\rm log_2(3)+\log_2 x=3\] like this OR yes right!!
that's how we should condense log equations \[\huge\rm log_2 (3 \times x)=3\] now we should convert log to exponential form
|dw:1439575720373:dw|
That just absolutely confused me.
well you will see how easy it's just have to move that variables around
right side term becomes exponent of base and move the left side term to the right side |dw:1439575981638:dw|
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.?
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
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|
here is the example to convert log to exponential form \[\huge\rm log_\color{ReD}{b} x = \color{blue}{y}\] b is base \[\huge\rm \color{ReD}{b}^\color{blue}{y} =x\]
\[3^{2} =4x.?\]
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!

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