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Logarithms Question

This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2
Reply 1
Avatar for AQASUX
AQASUX
OP
7
I'm guessing log(2^n - 10 doe not equal log2^n - 1
Reply 2
Avatar for Notnek
Notnek
21
Original post by AQASUX
This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2

What is the actual question? I think you missed something from your post.
Reply 3
Avatar for JT4NG56
JT4NG56
4
Original post by AQASUX
This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2


The question is unclear.... In the first line you wrote 7(2^n − 1) and in the first line of your working, you wrote log7(2n^-1). Also, where did you get the 200000 from? It might help if you post a screenshot or image of the question.
Reply 4
Avatar for AQASUX
AQASUX
OP
7
There's some previous parts where you eventually find the equation 7(2^n -1). It's a geometric progression actually not really an equation. It then asks for the inequality i've put above. It's off mei C2 june 2010 12(ii)
Original post by notnek
What is the actual question? I think you missed something from your post.
Reply 5
Avatar for Notnek
Notnek
21
Original post by AQASUX
There's some previous parts where you eventually find the equation 7(2^n -1). It's a geometric progression actually not really an equation. It then asks for the inequality i've put above. It's off mei C2 june 2010 12(ii)

7(2n−1)>200000\displaystyle 7(2^n-1)>200000

Rearrange to make n the subject, then you'll have something like this:

n>log(ab+c)\displaystyle n>log\left(\frac{a}{b}+c\right)

At this point I would recommend combining ab\frac{a}{b} with cc inside the log (adding fractions).

Post all your working if you're still stuck.

You missed the inequality out of your original post, which is why we couldn't help you initially.
(edited 8 years ago)
Reply 6
Avatar for JT4NG56
JT4NG56
4
Original post by AQASUX
There's some previous parts where you eventually find the equation 7(2^n -1). It's a geometric progression actually not really an equation. It then asks for the inequality i've put above. It's off mei C2 june 2010 12(ii)


Here is my solution:

Spoiler

Reply 7
Avatar for Notnek
Notnek
21
Original post by JT4NG56
Here is my solution:

Spoiler


Please don't post full solutions. It's against the rules of the maths forum.
Reply 8
Avatar for JT4NG56
JT4NG56
4
Original post by AQASUX
This is the equation 7(2^n − 1). It asks to show that n > (log 200 007 − log 7) / log 2. I'm just not sure how to get to log 200007 because when i work it out it's just 200000. I'm sure its something to do with manipulating the 1 but not sure how to do it. My current working is:

log7(2n^-1)>log200000
log7 + log 2^n - log1 > log200000
(log1=0) so log7+nlog2>log200000
nlog2>log200000 - log7
n>(log200000 - log7)/ log 2


Your error is between the first and second line:
log7(2^n -1) = log7 + log2^n -1
but log2^n -1 is not the same as log2^n - log1
Reply 9
Avatar for AQASUX
AQASUX
OP
7
Original post by JT4NG56
Here is my solution:

Spoiler


Perfect, thank you
Reply 10
Avatar for JT4NG56
JT4NG56
4
Original post by notnek
Please don't post full solutions. It's against the rules of the maths forum.


I have put the full solution inside a Spoiler and I also let him know where he went wrong in the post after it...
im so glad im not taking whatever the fu.k this is in college
(edited 8 years ago)
Reply 12
Avatar for JT4NG56
JT4NG56
4
Original post by valbrechts
im so glad im not taking whatever the fu.k this is in college


It's AS Maths.
Reply 13
Avatar for Notnek
Notnek
21
Original post by JT4NG56
I have put the full solution inside a Spoiler and I also let him know where he went wrong in the post after it...

It's still against the rules.

Giving the full solution was unnecessary. The OP could have worked it out themselves which is always more beneficial.
Reply 14
Avatar for AQASUX
AQASUX
OP
7
Original post by JT4NG56
Your error is between the first and second line:
log7(2^n -1) = log7 + log2^n -1
but log2^n -1 is not the same as log2^n - log1

Yeh I thought so but couldn't think of a way around it
Reply 15
Avatar for JT4NG56
JT4NG56
4
Original post by notnek
It's still against the rules.

Giving the full solution was unnecessary. The OP could have worked it out themselves which is always more beneficial.


Okay.

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