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C2 Logs Edexcel watch

1. Hi guys, I'm stuck on a few logs questions on the Mixed Exercise in the Logs Chapter in the C2 Book. Please may someone give me some help?

1) Find the possible values of for which 2^2x+1

and

3) Given that , express in terms of ,

a) ,
b) .

2. 2^(2x + 1) = 2^(2x) * 2^(1) by C1 rules, right?

= 2 * 2^(2x)

Can you see how to do the question now? If not, let u = 2^x and sub it in.

Post some working for question 3.
3. Thanks for the help, but I am still a bit confused on question 1 >_< can you explain a bit more? O_o

and on number 3, how do i start it? Do I sub log16 into log2?
4. (Original post by xsweetwhispers)
Thanks for the help, but I am still a bit confused on question 1 >_< can you explain a bit more? O_o
Well, what exactly is confusing you? Post your working and explanation of where + why you get stuck.

and on number 3, how do i start it? Do I sub log16 into log2?
Note that 16 = 2^4...
5. What are the inequality signs for?

Baiscally you can turn the first one into a quadratic equation. y=2^x can be used to put that into a quad as 2^2x becomes y^2 and then you end up with a y term to the single power later in the equation...solve for y and then sub back in for 2^x
6. On number one, this is how far I got...

2^(2x+1) = 3(2^x)-1
2^(2x) * 2^1 = 3(2^x)-1
2^(2x) * 2 = 3(2^x)-1

let y = 2^x

2 * y = 2^(2x) * 2 = 3(y)-1
2 * y = 2^(2x) * 2 = 3y-1

& now I'm back to stuck again >_<

----------------------------------

On number 3a), this is as far I've got to...

I think it is wrong though, can someone correct me?
7. (Original post by xsweetwhispers)
On number one, this is how far I got...

2^(2x+1) = 3(2^x)-1
2^(2x) * 2^1 = 3(2^x)-1
2^(2x) * 2 = 3(2^x)-1

let y = 2^x

2 * y = 2^(2x) * 2 = 3(y)-1
2 * y = 2^(2x) * 2 = 3y-1

& now I'm back to stuck again >_<

-

I don't think you have typed out the question correctly, but if you have then I think the answer is x=0

8. thanks for the help ^^ I got it now
but here goes another one I'm stuck on now...

quest 9.

If and , find and
9. (Original post by xsweetwhispers)
let y = 2^x

2 * y = 2^(2x) * 2 = 3(y)-1
You went wrong here.

2 * y = 2^x * 2, not 2^(2x) * 2.

So what is 2^(2x) * 2 in terms of y? Well, 2^(2x) = (2^x)^2 isn't it? So it's 2^(2x) * 2 = (2^x)^2 * 2 = y^2 * 2 = 2y^2.

Your solution to 3a doesn't make sense because you come up with x from nowhere and x isn't defined in the question.

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Updated: February 7, 2010
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