C2, More LOG problems

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.

1. Solve, giving your answer to 3 significant figures:
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

d). Calculate log2(beta), giving your answer to 3 significant figures.

Thank you for any help you can give.

Reply 1

Knogle

Question 1:

lgx/lg2 + lgx/lg4 = 2

lg4lgx + lg2lgx = 2lg2lg4
(lg4 + lg2)lgx = 2lg2lg4
lgx = 2lg2lg4 / (lg4 + lg2) = ...

Reply 2

meef cheese

JasonN

2)a) 3 + 2log2X = log2Y
=> 3 + log2X² = log2Y
=>2^( 3 + log2X²) = 2^(log2Y)
=> 2³.2^log2X² = y
=> y = 8x²

Reply 3

JasonN

e-unit

2)a) 3 + 2log2X = log2Y
=> 3 + log2X² = log2Y
=>2^( 3 + log2X²) = 2^(log2Y)
=> 2³.2^log2X² = 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.

Reply 4

Knogle

He just introduced the two, and made the existing terms the order of the value "2".

Did that make sense?

Reply 5

JasonN

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.

Reply 6

Knogle

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: x^log_xa = a.

Reply 7

JasonN

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: x^log_xa = a.

Right.

And the number 2 was introduced as the base in this question is 2?

Pretty much, yes.

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

Reply 10

The Orientalist

Knogle

lgx/lg2 + lgx/lg4 = 2

You meant lnx/ln2+lnx/ln4=2 right?

Reply 11

meef cheese

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

Reply 12

JasonN

Anyone had a go at parts b/c/d.

Just had another go and really am struggling.

Reply 13

The Orientalist

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.

Reply 14

Knogle

Pravin

You meant lnx/ln2+lnx/ln4=2 right?

Both are correct. ln is just log_e.

Reply 15

WilliamSlim

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

(edited 9 years ago)