Core 2 Log question Watch

Jimmy20002012
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#1
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#1
Could anyone help me out in the question below, don't know where I have gone wrong: it's part b

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Here is my solution:

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Thanks


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Numan786
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You were doing it right till you decided to factor out the 2 for the x ad 1 for some reason . you should have added the 1 to the other side and then divided it all by 2.

So it should be ((ln.05/ln3) + 1)/2

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Jimmy20002012
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(Original post by Numan786)
You were doing it right till you decided to factor out the 2 for the x ad 1 for some reason . you should have added the 1 to the other side and then divided it all by 2.

So it should be ((ln.05/ln3) + 1)/2

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So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


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Sataris
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(Original post by Jimmy20002012)
So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


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Well you don't have to, but when you divide 2x-1 by 2 you get x - \frac{1}{2} You divide every term by 2.
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TenOfThem
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(Original post by Jimmy20002012)
So do you always get rid of everything else, and then what x is divided by you take to the other side to find x?


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Can you solve 2x-1=5
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Jimmy20002012
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(Original post by TenOfThem)
Can you solve 2x-1=5
2x= 6
X=6/2= 3



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TenOfThem
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(Original post by Jimmy20002012)
2x= 6
X=6/2= 3
So yours was 2x-1 = something

SO you still have to +1 and then /2

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Jimmy20002012
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(Original post by TenOfThem)
So yours was 2x-1 = something

SO you still have to +1 and then /2

Thanks for the help


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Jimmy20002012
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(Original post by TenOfThem)
So yours was 2x-1 = something

SO you still have to +1 and then /2

Could you also help me in this sequences question:
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Surely if the 21st term is twice greater than the 16th terms, the equation applied to this question would be 2 x the 21st term??


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TenOfThem
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#10
(Original post by Jimmy20002012)
Could you also help me in this sequences question:
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Name:  ImageUploadedByStudent Room1399660422.157947.jpg
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Surely if the 21st term is twice greater than the 16th terms, the equation applied to this question would be 2 x the 21st term??
Why do you think that?

a is twice b means a=2b
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JoelHopkins
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Jimmy20002012
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#12
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#12
If you had a question like

2 x 3^n-1 > 4 x 10^15 and you had to solve for n,

Would you apply logs and get

(log4 + log10^15 / log2 + log3) +1 (Log base 10 used)

Is this right or another way of doing it?? Thanks


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lazy_fish
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#13
(Original post by Jimmy20002012)
...
Method looks fine but see a mistake.

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Jimmy20002012
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(Original post by lazy_fish)
Method looks fine but see a mistake.

What's the mistake??


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lazy_fish
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#15
(Original post by Jimmy20002012)
...

Just to clarify, your answer was  \displaystyle n > \frac {\log 4 + \log 10^{15}} {\log 2 + \log 3} +1

If so, there is a mistake somewhere.

Hope that helps.
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Jimmy20002012
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(Original post by lazy_fish)

Just to clarify, your answer was  \displaystyle n > \frac {\log 4 + \log 10^{15}} {\log 2 + \log 3} +1

If so, there is a mistake somewhere.

Hope that helps.
Actually can't see he mistake


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lazy_fish
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#17
(Original post by Jimmy20002012)
...
I mean mistake reaching that answer.

Mine was  \displaystyle n > \frac {\log 2 + 15} {\log 3} +1

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Jimmy20002012
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(Original post by lazy_fish)
I mean mistake reaching that answer.

Mine was  \displaystyle n > \frac {\log 2 + 15} {\log 3} +1

Why is it that


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lazy_fish
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#19
(Original post by Jimmy20002012)
...
Maybe show the working?
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