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# C2, More LOG problems watch

1. Logs is the only thing on C2 i'm really struggling on. Haven't done them for a while and don't know how to go about answering the questions where different bases are used, and where just a number is involved. Here are two examples. I would really appreciate some help.

log2X + log4X = 2

2a). Given that 3 + 2log2X = log2Y, show that Y=8(x)squared.

b). Hence, or otherwise, find the roots (alpha sign) and (beta sign) where (alpha) < (beta), of the equation:
3 + 2log2X = log2(14x - 3).

c). Show that log2(alpha) = -2

2. Question 1:

lgx/lg2 + lgx/lg4 = 2

lg4lgx + lg2lgx = 2lg2lg4
(lg4 + lg2)lgx = 2lg2lg4
lgx = 2lg2lg4 / (lg4 + lg2) = ...
3. (Original post by JasonN)

Logs is the only thing on C2 i'm really struggling on. Haven't done them for a while and don't know how to go about answering the questions where different bases are used, and where just a number is involved. Here are two examples. I would really appreciate some help.

log2X + log4X = 2

2a). Given that 3 + 2log2X = log2Y, show that Y=8(x)squared.

b). Hence, or otherwise, find the roots (alpha sign) and (beta sign) where (alpha) < (beta), of the equation:
3 + 2log2X = log2(14x - 3).

c). Show that log2(alpha) = -2

2)a) 3 + 2log2X = log2Y
=> 3 + log2X2 = log2Y
=>2^( 3 + log2X2) = 2^(log2Y)
=> 23.2log2X2 = y
=> y = 8x2
4. (Original post by e-unit)
2)a) 3 + 2log2X = log2Y
=> 3 + log2X2 = log2Y
=>2^( 3 + log2X2) = 2^(log2Y)
=> 23.2log2X2 = y
=> y = 8x2
Thank you for your reply, but can you please explain for me how you got to 2^( 3 + log2X2) = 2^(log2Y) I don't understand.

and how you got from that to the answer.
5. He just introduced the two, and made the existing terms the order of the value "2".

Did that make sense?
6. (Original post by Knogle)
He just introduced the two, and made the existing terms the order of the value "2".

Did that make sense?
I understand what you mean, i just don't understand why that was done.

What rule is that?
And why does 2^(log2Y) = y

Sorry for my lack of understanding, just haven't been taught it for a while.
7. (Original post by JasonN)
I understand what you mean, i just don't understand why that was done.

What rule is that?
And why does 2^(log2Y) = y

Sorry for my lack of understanding, just haven't been taught it for a while.
That isn't a rule per se, it's just something he introduced to solve the equation.

And this is a general rule of thumb: xlogxa = a.
8. (Original post by Knogle)
That isn't a rule per se, it's just something he introduced to solve the equation.

And this is a general rule of thumb: xlogxa = a.
Right.

And the number 2 was introduced as the base in this question is 2?
9. Pretty much, yes.
10. You can save time by also remembering loga - logb = log(a/b).
in Q1 and Q2 apply this.
So Q2 would be 3= log2(y/x^2) hence 2^3= y/x^2
11. (Original post by Knogle)
lgx/lg2 + lgx/lg4 = 2
You meant lnx/ln2+lnx/ln4=2 right?
12. (Original post by Pravin)
You meant lnx/ln2+lnx/ln4=2 right?
They're both correct.

Actually, you can take logs of any base you like.

ln = base e and lg = base 10
13. Anyone had a go at parts b/c/d.

Just had another go and really am struggling.
14. For b, solve the quadratic equation 8x^2-14x+3=0 The solutions represent alpha and beta where alpha smaller than beta. I'm sure you can do c and d after this, good luck.
15. (Original post by Pravin)
You meant lnx/ln2+lnx/ln4=2 right?
Both are correct. ln is just loge.
16. (Original post by Knogle)
Question 1:
I'm stuck on this question and the above method of working out makes no sense to me

atm all i know is to do

logX/log2 + logX/log4=2
and then

logX*log4 + log2*logX = 2

But after that i am lost

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