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Could anyone explain this to me? watch

1. Ok here's the question

log(base2)x=12 so find the following:
A) log(base2)x3
B)log(base2)16x
C)log(base2)rootx

Im struggling on how to start and what is actually being asked of me.

Do I need to find x? Is there a particular rule of logs I should be using?

I've got some brilliant help on TSR in the past so hoping you guys can help me again on this!

2. Just use these laws of logarithms:

3. first log(base2)x=12 means 2^12 = x
so for A, using the definition again 2^(something) =x^3 now it should be clear then how to proceed with the rest
4. (Original post by Mr M)
Just use these laws of logarithms:

Okay, so for

This means ?

And

Means
If so, then what? I don't understand what the acutal answer is supposed to me? And which rule do I use for the square root of x one?
(Sorry for the continuous questions!)
5. (Original post by breakeven)
Okay, so for

This means ?

And

Means
If so, then what? I don't understand what the acutal answer is supposed to me?
So you are very close now.

What does (massive hint, you are told the answer in the question) ?
6. (Original post by Ayakashi)
first log(base2)x=12 means 2^12 = x
so for A, using the definition again 2^(something) =x^3 now it should be clear then how to proceed with the rest
So, x=4096?

and Log(base2)x3=36?
Is 36 the actual final answer?
7. (Original post by Mr M)
So you are very close now.

What does (massive hint, you are told the answer in the question) ?
I got a) = 36
b) = 16
and c) = 6

Is that right?
8. (Original post by breakeven)
I got a) = 36
b) = 16
and c) = 6

Is that right?
Yes.
9. (Original post by breakeven)
So, x=4096?

and Log(base2)x3=36?
Is 36 the actual final answer?
I wouldn't do this. It is correct but unnecessarily complicated.
10. (Original post by Mr M)
Yes.
Thanks a million for your help and patience, you rock!
11. (Original post by breakeven)
Thanks a million for your help and patience, you rock!
12. (Original post by Mr M)
Last question on my sheet and I'm stuck again

Wondered if you might want to help out again, its fine if you don't

f(x)=x3-x2-7x+c
f(4)=0
a) Find C (which I got -20 for)
b) Factorise f(x) as the product of a linear factor and a quadratic factor
c) hence show that, apart from x=4, there are no real values of x for which f(x)=0

B+C are confusing me.
I guessed that I needed to find the quotient using (x-4) and then factorise it using the quadratic formula which didnt work.

Not sure what it's asking me to do now:/
13. (Original post by breakeven)
Last question on my sheet and I'm stuck again

Wondered if you might want to help out again, its fine if you don't

f(x)=x3-x2-7x+c
f(4)=0
a) Find C (which I got -20 for)
b) Factorise f(x) as the product of a linear factor and a quadratic factor
c) hence show that, apart from x=4, there are no real values of x for which f(x)=0

B+C are confusing me.
I guessed that I needed to find the quotient using (x-4) and then factorise it using the quadratic formula which didnt work.

Not sure what it's asking me to do now:/
You know that f(4)=0 so (x-4) is a factor (from factor theorem).

You should be able to write down the values of A and C immediately and figure out B (expand the brackets and equate coefficients if you need to).
14. (Original post by Mr M)
You know that f(4)=0 so (x-4) is a factor (from factor theorem).

You should be able to write down the values of A and C immediately and figure out B (expand the brackets and equate coefficients if you need to).
So, (x-4)(x2+3x+5)?

A=1 B=3 C=5?

And for part c? xx
15. (Original post by breakeven)
And for part c? xx
What method do you know that shows a quadratic has no real roots?
16. (Original post by Mr M)
What method do you know that shows a quadratic has no real roots?
Do you mean show that the discriminant is less than 0?
17. (Original post by breakeven)
Do you mean show that the discriminant is less than 0?
18. (Original post by Mr M)
I'm sending you lots of air hugs right now! I appreciate your help and explanations, they really do help so much
19. (Original post by breakeven)
I'm sending you lots of air hugs right now! I appreciate your help and explanations, they really do help so much
Ok - you are welcome.

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Updated: April 14, 2013
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