anonymous
  • anonymous
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???
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Nnesha
  • Nnesha
what's the original question ?
anonymous
  • anonymous
Log\/2(3)+log\/2(x)=3
Nnesha
  • Nnesha
i mean equation \[\log_2(3)+\log_2 x=3\] like this OR

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Nnesha
  • Nnesha
\[\frac{ \log_2 3 }{ \log_2 x} =3\]
Nnesha
  • Nnesha
maybe or maybe division hmm which one is correct ? graysongraddyLOL
anonymous
  • anonymous
the first one is the original
Nnesha
  • Nnesha
oh okay familiar with the log properties ??
anonymous
  • anonymous
Not exactly.
Nnesha
  • 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
  • Nnesha
look at that and tell me which one you should apply ?
anonymous
  • anonymous
The last one.?
anonymous
  • anonymous
Math just absolutely dumbfounds me.
Nnesha
  • Nnesha
hmm alright there is PLUS sign between log now look at the log identities
anonymous
  • anonymous
the subexponent of 2 is the identity correct.?
Nnesha
  • Nnesha
2 is base
Nnesha
  • 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
  • anonymous
so with the equation that i am using, How do i get a Log on the other side of the equal sign.?
Nnesha
  • 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
  • anonymous
So multiply 3 and x.?
Nnesha
  • Nnesha
\[\huge\rm log_2(3)+\log_2 x=3\] like this OR yes right!!
Nnesha
  • 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
  • Nnesha
|dw:1439575720373:dw|
anonymous
  • anonymous
That just absolutely confused me.
Nnesha
  • Nnesha
well you will see how easy it's just have to move that variables around
Nnesha
  • Nnesha
right side term becomes exponent of base and move the left side term to the right side |dw:1439575981638:dw|
anonymous
  • anonymous
so log2 (3) = 3x.?
Nnesha
  • Nnesha
we need to convert to exponential there shouldn't be an log b=2 how would write 2 with 3 exponent ?
Nnesha
  • Nnesha
2 to the 3 ppower
anonymous
  • anonymous
so 2^3=3?
Nnesha
  • Nnesha
2^3 = 3x not just x
Nnesha
  • Nnesha
now solve for x
anonymous
  • anonymous
\[x=\frac{ 8 }{ 3 }\]
Nnesha
  • Nnesha
:)
anonymous
  • anonymous
Okay cool :)
anonymous
  • anonymous
Can you give me a example logarithm so i can see if i got the hang of it.?
Nnesha
  • Nnesha
why not!!
anonymous
  • anonymous
Thank you so much :D
Nnesha
  • Nnesha
\[\huge\rm log_3 (4) - \log_3 (y) = 2\] look at the log properties
anonymous
  • anonymous
So id use the first one right.?
Nnesha
  • Nnesha
yea quotient property
anonymous
  • anonymous
then it would go to \[\log_{3}(4 \div x)=3\]
Nnesha
  • Nnesha
yep right
Nnesha
  • Nnesha
it's y not x but it's okay :P
Nnesha
  • Nnesha
i'm not gonna take off points like teachers do lol
anonymous
  • anonymous
Oh whoops lol
anonymous
  • anonymous
So it would go to \[3^{4}=3x\]
Nnesha
  • Nnesha
hmm no
anonymous
  • anonymous
Where did i go wrong.?
Nnesha
  • 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
  • Nnesha
try again!
anonymous
  • anonymous
okay so it would go to \[\frac{ 4 }{ x } = 2x \]
Nnesha
  • Nnesha
how did you get two x's ?
anonymous
  • anonymous
wait it would just be 2 on the right side
anonymous
  • anonymous
then i multiply 4 from the left side to get 8 on the right side so x=8.?
Nnesha
  • Nnesha
|dw:1439577454761:dw|
Nnesha
  • 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
  • anonymous
\[3^{2} =4x.?\]
Nnesha
  • 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
  • anonymous
\[x=\frac{ 4 }{ 9 }\]
Nnesha
  • Nnesha
yep!
anonymous
  • anonymous
Thank you so much!
Nnesha
  • Nnesha
my pleasure!

Looking for something else?

Not the answer you are looking for? Search for more explanations.