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Could anyone explain this to me?

Ok here's the question:smile:

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!

Thanks in advance :smile:
Just use these laws of logarithms:

log2(ab)=log2a+log2b\log_2 (ab) = \log_2 a + \log_2 b

log2an=nlog2a\log_2 a^n = n \log_2 a
Reply 2
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
Reply 3
Original post by Mr M
Just use these laws of logarithms:

log2(ab)=log2a+log2b\log_2 (ab) = \log_2 a + \log_2 b

log2an=nlog2a\log_2 a^n = n \log_2 a


Okay, so for log2(ab)=log2a+log2b\log_2 (ab) = \log_2 a + \log_2 b

This means log2(16x)=log216+log2x\log_2 (16x) = \log_2 16 + \log_2 x?

And log2an=nlog2a\log_2 a^n = n \log_2 a

Means log2x3=3log2x\log_2 x^3 = 3 \log_2 x
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!)
(edited 10 years ago)
Original post by breakeven
Okay, so for log2(ab)=log2a+log2b\log_2 (ab) = \log_2 a + \log_2 b

This means log2(16x)=log216+log2x\log_2 (16x) = \log_2 16 + \log_2 x?

And log2an=nlog2a\log_2 a^n = n \log_2 a

Means log2x3=3log2x\log_2 x^3 = 3 \log_2 x
If so, then what? I don't understand what the acutal answer is supposed to me?


So you are very close now.

What does log2x=\log_2 x = (massive hint, you are told the answer in the question) ?
Reply 5
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?
Reply 6
Original post by Mr M
So you are very close now.

What does log2x=\log_2 x = (massive hint, you are told the answer in the question) ?


I got a) = 36
b) = 16
and c) = 6

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

Is that right?


Yes.
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.
Reply 9
Original post by Mr M
Yes.


Thanks a million for your help and patience, you rock! :smile:
Original post by breakeven
Thanks a million for your help and patience, you rock! :smile:


:rock:
Reply 11


Last question on my sheet and I'm stuck again:frown:

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

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:/
Original post by breakeven
Last question on my sheet and I'm stuck again:frown:

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

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).

f(x)=(x4)(Ax2+Bx+C)f(x)=(x-4)(Ax^2+Bx+C)

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).
Reply 13
Original post by Mr M
You know that f(4)=0 so (x-4) is a factor (from factor theorem).

f(x)=(x4)(Ax2+Bx+C)f(x)=(x-4)(Ax^2+Bx+C)

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
Original post by breakeven
And for part c? xx


What method do you know that shows a quadratic has no real roots?
Reply 15
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?
Original post by breakeven
Do you mean show that the discriminant is less than 0?


:yes:
Reply 17


I'm sending you lots of air hugs right now! I appreciate your help and explanations, they really do help so much :biggrin:
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 :biggrin:


Ok - you are welcome.

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