Natural Logarithms

Question is express (112 + 6ln4 - 6ln8)÷4 in the form a + bln2.

This question has spun me out for the last half hour, please help before I explode.

Reply 1

Christophicus

GarageMc

Question is express (112 + 6ln4 - 6ln8)÷4 in the form a + bln2.

This question has spun me out for the last half hour, please help before I explode.

(112 + 6ln4 - 6ln8)/4
= 112/4 + (6ln4-6ln8)/4
= 28 + [6ln2²-6ln2³]/4
= 28 + [12ln2-18ln2]/4
= 28 + [-6ln2]/4
= 28-1.5ln2

Reply 2

Dimez

(112 + 6ln4 - 6ln8)/4

(112+ 12ln2 - 18ln2)/4

(112 +[12ln2 -18ln2]) /4

(112 - 6ln2)/4

28 -3/2[ln2]

Reply 3

manps

Dimez

(112 + 6ln4 - 6ln8)/4

(112+ 12ln2 - 18ln2)/4

(112 +[12ln2/18ln2]) /4

28 + 1/6[ln2]

--------------

****, you're right WM, I used that silly logarithm law.

Damn you!

This is wrong!
Correct answer is 28 - 3/2ln2

Working,

[112 + ln(4^6/8^6)]/4

[112 + ln(4^6/2^6.4^6)]/4

[112 + ln(1/2^6)]/4

[112 + ln2^-6]/4

[112 - 6ln2]/4

∴ 28 - 3/2ln2

Reply 4

Christophicus

GarageMc

Question is express (112 + 6ln4 - 6ln8)÷4 in the form a + bln2.

This question has spun me out for the last half hour, please help before I explode.

(112 + 6ln4 - 6ln8)/4
= 112/4 + (6ln4-6ln8)/4
= 28 + [6ln2²-6ln2³]/4
= 28 + [12ln2-18ln2]/4
= 28 + [-6ln2]/4
= 28-1.5ln2

lol

Reply 5

Dimez

manps

This is wrong!
Correct answer is 28 - 3/2ln2

Wha Wha? What you on about? :p:

Reply 6

manps

Dimez

Wha Wha? What you on about? :p:

oh dear me, how some love the edit button :p:

Reply 7

Dimez

manps

oh dear me, how some love the edit button :p:

Tell me about. :rolleyes:

You edited your post, for all we know, you could have copied my working out! :p:

Reply 8

GarageMc

Thanks guys! The book never tells you that 6ln4 can also equal 12ln2! Yet they ask a question on it!

Reply 9

Dimez

GarageMc

Thanks guys! The book never tells you that 6ln4 can also equal 12ln2! Yet they ask a question on it!

It's the power law. The book dosen't have to, you have to deduce it.

Reply 10

Christophicus

GarageMc

Thanks guys! The book never tells you that 6ln4 can also equal 12ln2! Yet they ask a question on it!

Hmm, I take it you're new to logs. Well if you want I'll give you the main rules.

logab = loga + logb
log(a/b) = loga - logb
log_ab = log_tb/log_ta
log(1/a) = -loga

If y = x^a
then log_xy = a

If f = x^a and g = x^b
fg = x^a+b
logfg = (a+b)logx
logfg = alogx + blogx
logfg = logx^a + logx^b
logfg = logf + logg

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