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C2 edexcel log question

User940459

help i cant answer this at all

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Reply 1

Notnek

What is 3 written as a base 2 logarithm? Then combine the logs on the right hand side and rewrite the log on the left hand side. You should end up with something like:

\log_2 A = \log_2 B

And then you can say that A=B.

Reply 2

raheem94

Original post by Tishax2

help i cant answer this at all

\displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3

Now use the rule,

\displaystyle log_ab-log_ac = log_a\left(\frac{b}{c}\right)

Reply 3

User940459

Original post by raheem94

\displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3

Now use the rule,

\displaystyle log_ab-log_ac = log_a\left(\frac{b}{c}\right)

what do i do from there :s-smilie:

Reply 4

User940459

Original post by notnek

What is 3 written as a base 2 logarithm? Then combine the logs on the right hand side and rewrite the log on the left hand side. You should end up with something like:

\log_2 A = \log_2 B

And then you can say that A=B.

3 is jsut a number

and y >0

Reply 5

Notnek

Original post by Tishax2

3 is jsut a number

and y >0

3 is just a number and so is

log_2 2^3

which is equal to 3 (y>0 doesn't change this).

So you have:

\displaystyle 2\log_2 y = \log_2 8 + \log_2 (y+6)

Can you follow the next steps in my last post?

Also, it's a good idea for you to try both methods that have been given to you in this thread so you can decide which one you are more comfortable with.

(edited 11 years ago)

Reply 6

aeyurttaser13

2log2 y - log2 y+6 =3
2log2 y/(y+6) =3
log 2 (y/(y+6)squared = 3
2 cubed = (y/(y+6)) squared.. the rest u should be able to work out :tongue:

Reply 7

Notnek

Original post by aeyurttaser13

2log2 y - log2 y+6 =3
2log2 y/(y+6) =3
log 2 (y/(y+6)squared = 3
2 cubed = (y/(y+6)) squared.. the rest u should be able to work out :tongue:

You've made a mistake in your second line although it's hard to know for sure because your working isn't clear.

You cannot combine two logs which have different coefficients. In general,

\displaystyle 2\log_c a - \log_c b \neq 2\log_c \frac{a}{b}

(edited 11 years ago)

Reply 8

User940459

Original post by notnek

3 is just a number and so is

log_2 2^3

which is equal to 3 (y>0 doesn't change this).

So you have:

\displaystyle 2\log_2 y = \log_2 8 + \log_2 (y+6)

im sorry i really do not understand your method

and especically why you put 3 as a power when it says 3+...

thanks

Reply 9

raheem94

Original post by Tishax2

what do i do from there :s-smilie:

\displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3 \implies log_2\left(\frac{y^2}{y+6}\right)=3

Now use the rule,

log_ab=x \implies a^x = b

You will get a quadratic so you will obtain 2 solutions, reject the value which is less than zero.

Reply 10

raheem94

Original post by Tishax2

im sorry i really do not understand your method

and especically why you put 3 as a power when it says 3+...

thanks

Remember,

3 = log_22^3 = 3log_22

Reply 11

Notnek

Original post by Tishax2

im sorry i really do not understand your method

and especically why you put 3 as a power when it says 3+...

thanks

No problem. I'll try to explain it further.

What I have done is say that 3 is equal to

log_2 8

. In your first lessons on logarithms, you should have learnt why this is true. Put

log_2 8

into your calculator to convince yourself if you're still not sure.

Remember,

log_2 2^1 = 1, log_2 2^2 = 2, log_2 2^3 = 3

etc.

Next I have just replaced 3 with

log_2 8

, since they are the same thing.

Reply 12

aeyurttaser13

Original post by notnek

You've made a mistake in your second line although it's hard to know for sure because your working isn't clear.

You cannot combine two logs which have different coefficients. In general,

\displaystyle 2\log_c a - \log_c b \neq 2\log_c \frac{a}{b}

oops, ib math exam tmrw and still stupid mistakes! ughh thx for pointing it out its like a warning before my exam tmrw that i gotta DOUBLE CHECK haha :tongue:

shouldve been log_c a squared - log_c b = log_c a (squared) / b :smile:

Reply 13

User940459

Original post by raheem94

\displaystyle 2log_2y=3+log_2(y+6) \implies log_2y^2 - log_2(y+6) =3 \implies log_2\left(\frac{y^2}{y+6}\right)=3

Now use the rule,

log_ab=x \implies a^x = b

You will get a quadratic so you will obtain 2 solutions, reject the value which is less than zero.

how do i get the quad equation from here :s

Reply 14

User940459

Original post by notnek

No problem. I'll try to explain it further.

What I have done is say that 3 is equal to

log_2 8

. In your first lessons on logarithms, you should have learnt why this is true. Put

log_2 8

into your calculator to convince yourself if you're still not sure.

Remember,

log_2 2^1 = 1, log_2 2^2 = 2, log_2 2^3 = 3

etc.

Next I have just replaced 3 with

log_2 8

, since they are the same thing.

thanks but we did not learn that in class :s

Reply 15

Jay™

Original post by Tishax2

how do i get the quad equation from here :s

y^2 / (y+6) = 2^3

y^2 / (y+6) = 8

y^2 = 8(y+6)

y^2 = 8y + 48

y^2 - 8y - 48 = 0

And solve.

Reply 16

Notnek

Original post by Tishax2

how do i get the quad equation from here :s

\displaystyle \log_2 \frac{y^2}{y+6} = 3 \implies \frac{y^2}{y+6}=2^3

If this confuses you then what's happening is similar to e.g this:

\displaystyle \log_2 x = 5 \implies x=2^5

(5 is an arbitrary number that I picked out of the air to show you an example)

Do you understand this? If not, you may need to revise some basic logarithms work.