# Heinemann Core 3 Book: ExponentialWatch

This discussion is closed.
#1
rite 1st two bits are easy and then where they did they pull the 3rd bit from...it must have obviously to do with the easy parts os any help appreciated...

f(x) = 3x^3 - 4x^2 - 5x + 2

a) show (x+1) is a factor of f(x)
b) factorise f(x) completely - which gives (x+1)(3x-1)(x-2) as our answer
c) solve giving answers to 2dp the equation:

3[ln(2x)]^3 - 4[ln(2x)]^2 - 5ln(2x) = 0 x>0

rite so obviously im stuck on part c
0
13 years ago
#2
(Original post by nacero2000)
rite 1st two bits are easy and then where they did they pull the 3rd bit from...it must have obviously to do with the easy parts os any help appreciated...

f(x) = 3x^3 - 4x^2 - 5x + 2

a) show (x+1) is a factor of f(x)
b) factorise f(x) completely - which gives (x+1)(3x-1)(x-2) as our answer
c) solve giving answers to 2dp the equation:

3[ln(2x)]^3 - 4[ln(2x)]^2 - 5ln(2x) = 0 x>0

rite so obviously im stuck on part c
Solving for ln(2x) we get (from your previous factorisation)

ln2x = -1 (not possible)
ln2x = 1/3 -> x = exp(1/3)/2
ln2x = 2 -> x = exp(2)/2
0
13 years ago
#3
(Original post by nacero2000)
rite 1st two bits are easy and then where they did they pull the 3rd bit from...it must have obviously to do with the easy parts os any help appreciated...

f(x) = 3x^3 - 4x^2 - 5x + 2

a) show (x+1) is a factor of f(x)
b) factorise f(x) completely - which gives (x+1)(3x-1)(x-2) as our answer
c) solve giving answers to 2dp the equation:

3[ln(2x)]^3 - 4[ln(2x)]^2 - 5ln(2x) = 0 x>0

rite so obviously im stuck on part c
rewriting u as ln2x, your eqn becomes

3u^3 - 4u² - 5u = 0
u(3u² - 4u - 5) = 0
u = 0,
u = {4 +/- rt(16 + 60)} / 6

u = 0, 2.1196, -0.7863
ln2x = 0, 2.1196, -0.7863
x = 0.5, 4.1639, 0.2278
x = 0.50, 4.16, 0.23
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