# C2, More LOG problemsWatch

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#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

0
13 years ago
#2
Question 1:

lgx/lg2 + lgx/lg4 = 2

lg4lgx + lg2lgx = 2lg2lg4
(lg4 + lg2)lgx = 2lg2lg4
lgx = 2lg2lg4 / (lg4 + lg2) = ...
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13 years ago
#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
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#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.
0
13 years ago
#5
He just introduced the two, and made the existing terms the order of the value "2".

Did that make sense?
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#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.
0
13 years ago
#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.
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#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?
0
13 years ago
#9
Pretty much, yes.
0
13 years ago
#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
0
13 years ago
#11
(Original post by Knogle)
lgx/lg2 + lgx/lg4 = 2
You meant lnx/ln2+lnx/ln4=2 right?
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13 years ago
#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
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#13
Anyone had a go at parts b/c/d.

Just had another go and really am struggling.
0
13 years ago
#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.
0
13 years ago
#15
(Original post by Pravin)
You meant lnx/ln2+lnx/ln4=2 right?
Both are correct. ln is just loge.
0
4 years ago
#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
0
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