# Core 2 Log questionWatch

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

Here is my solution:

Thanks

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4 years ago
#2
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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#3
(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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4 years ago
#4
(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 by 2 you get You divide every term by 2.
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4 years ago
#5
(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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#6
(Original post by TenOfThem)
Can you solve 2x-1=5
2x= 6
X=6/2= 3

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4 years ago
#7
(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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#8
(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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#9
(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:

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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4 years ago
#10
(Original post by Jimmy20002012)
Could you also help me in this sequences question:

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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4 years ago
#11
0
#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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4 years ago
#13
(Original post by Jimmy20002012)
...
Method looks fine but see a mistake.

0
#14
(Original post by lazy_fish)
Method looks fine but see a mistake.

What's the mistake??

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4 years ago
#15
(Original post by Jimmy20002012)
...

If so, there is a mistake somewhere.

Hope that helps.
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#16
(Original post by lazy_fish)

If so, there is a mistake somewhere.

Hope that helps.
Actually can't see he mistake

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4 years ago
#17
(Original post by Jimmy20002012)
...
I mean mistake reaching that answer.

Mine was

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#18
(Original post by lazy_fish)
I mean mistake reaching that answer.

Mine was

Why is it that

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4 years ago
#19
(Original post by Jimmy20002012)
...
Maybe show the working?
0
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