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Pure 2 maths

jonnymcc2003

God I hate logarithms. I just can't get my head round them.

Given that .... 3 + 2log2x = log2y, show that y = 8x2

Any help? cheers.

Reply 1

4Ed

15

jonnymcc2003

God I hate logarithms. I just can't get my head round them.

Given that .... 3 + 2log2x = log2y, show that y = 8x2

Any help? cheers.

do you mean that y = 8 x^2 (squared)?

3 + 2log2 x = log2 y

3 = log2 y - log2 (x)^2
= log2 (y/x^2)

raise 2 by the power of both of them...

2^3 = 2^(log2 (y/x^2))

8 = y/x^2
y = 8x^2

Reply 2

jonnymcc2003

OP

Ahh great thanks. A couple more if you would be so kind....

solve the equation... log2x + 4logx2 = 5

if... xy = 64 and logxy + logyx = 5/2, find x and y

Cheers.

Reply 3

Fermat

8

loga b means the log of a to the base b.

You uase the relationship,

loga b * logb a = 1
============

a)
log2 x + 4 logx 2 =5
log2 x + 4/(log2 x) = 5

put u=log2 x,

u + 4/u = 5
u² +4 = 5u
u² -5u +4 = 0
(u-4)(u-1) = 0
u=4 or u=1
=======

log2 x = 4 => x=2^4 => x=16
log2 x = 1 => x=2^1 => x=2
===================

b)
logx y + logy x = 5/2
logx y + 1/(logy x) = 5/2

put u = logx y,

u + 1/u = 5/2
u² + 1 = 5u/2
2u² + 2 = 5u
2u² - 5u + 2 = 0
(2u-1)(u-2)=0
u=½ or u=2
=======

logx y = ½ =>y=x^(½) =>xy=x^(3/2)

now,

xy=64

or,

xy=2^6

therefore

x^(3/2) = 2^6
x=2^4
x=16
===

logx y = 2 =>y=x² =>xy=x³

now,

xy=64

or,

xy=2^6

therefore

x³ = 2^6
x=2²
x=4
===

Reply 4

jonnymcc2003

OP

Blooming heck. Thanks, although you wouldn't mind two more would you..

a. lgx + lgy = 1000
lg(3x + y) = 1

b. logy 2 = log4 32

Reply 5

m:)ckel

14

jonnymcc2003

Blooming heck. Thanks, although you wouldn't mind two more would you..

a. lgx + lgy = 1000
lg(3x + y) = 1

is this the one at the back of the edexcel P2 book? cos if it is, then there's a mistake in the question

Reply 6

jonnymcc2003

OP

mockel

is this the one at the back of the edexcel P2 book? cos if it is, then there's a mistake in the question

Yeah?

Reply 7

Fermat

8

a. lgx + lgy = 1000
lg(3x + y) = 1

I think there's a mistake here.

lg is the log to the base 2.

Now if lgx + lgy = 1000, then both x and y must be pretty large (on average)
For example if x=y=1024, then lgx = lgy = 10, giving lgx + lgy = 20 only.
But since lg(3x+y)=1, this gives 3x+y=2. And since both x and y must be positive (you can't take the log of a negative number) then they must both be small in order to add up to only 2!
Which contradicts what was supposed earlier about them being large.

Anyway, part b)

logy 2 = log4 32

put logy 2 = n, then

2=y^n
====

also,

log32 4 = n
4=32^n
2^2 = (2^5)^n=2^(5n)

equating coefficients,

2 = 5n
n=5/2
====

therefore,

2=y^(5/2)
2^2 = y^5
y=4^(1/5)
======

Reply 8

Barny

17

Yeh a) is a mistake, its in the back of the p2 book as part of a the practice paper. Its wrong, its supposed to be lg3 or 10 I think? something like that

Reply 9

jonnymcc2003

OP

Fermat

2 = 5n
n=5/2

QUOTE]

Is this part correct? The final answer's right though.

Pure 2 maths

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