C3 exponential equation

Jordan7

can you help me finish this qs i rli am stuck.

Solve the equation:
ln(x^2 +7x +12) - 1 = ln(2x^2 +9x +4)
giving answer in terms of e

if u raise it all to e you get:
(x^2 +7x +12) -e = (2x^2 +9x +4)
-e=(x^2 +2x -8)
-e=(x-2)(x+4)

what do i do now?

thanks

Nah that's wrong mate.

ln(x²+7x+12) - ln(2x²+9x+4) = 1 = lne
ln[(x²+7x+12)/((2x²+9x+4)] = lne
(x²+7x+12)/((2x²+9x+4) = e
(x²+7x+12) = e(2x²+9x+4)
(2e-1)x² + (9e-7)x + (4e-12) = 0

Can you solve that now?

Reply 3

OP

I have got no idea how to solve that, cheers foe gettin me to ther tho

Reply 4

OP

hang on il give it a go

Reply 5

steve10

7

It's just a quadratic in x.

Use the quadratic formula.

Reply 6

Jordan7

I have got no idea how to solve that, cheers foe gettin me to ther tho

It's a quadratic of the form Ax² + Bx + C = 0

Use

x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}

Reply 7

Fade Into Black

Widowmaker

Nah that's wrong mate.

ln(x²+7x+12) - ln(2x²+9x+4) = 1 = lne
ln[(x²+7x+12)/((2x²+9x+4)] = lne
(x²+7x+12)/((2x²+9x+4) = e
(x²+7x+12) = e(2x²+9x+4)
(2e-1)x² + (9e-7)x + (4e-12) = 0

Can you solve that now?

Perhaps what make it easier is factorising (x2+7x+12) = e(2x2+9x+4) at this line here to get (x+4)(x+3) = e(2x+1)(x+4) and cancelling down from there instead of using the quad. formula as it looks a bit crude and meticulous....
Hopefully that will make it easier for you Jordan 7.

Reply 8

OP

the questions says giv in terms of e.. does tht mean dont use a calculator?

Reply 9

Jordan7

the questions says giv in terms of e.. does tht mean dont use a calculator?

No, don't use a calculator. Just leave your answer in terms of e.

Reply 10

OP

if u substitute value for a b and c into the equation u get a rediculous value under the square root and without using a calculator its nr on impossible to find its root. Is ther not a simpler way to do this question because ther are onli 4 marks for it. Also the previous qs asks u to simplifiy (x^2 +7x+12)/(2x^2+9x+4) - does this not hav nething to do with it?

Reply 11

Jordan7

if u substitute value for a b and c into the equation u get a rediculous value under the square root and without using a calculator its nr on impossible to find its root. Is ther not a simpler way to do this question because ther are onli 4 marks for it. Also the previous qs asks u to simplifiy (x^2 +7x+12)/(2x^2+9x+4) - does this not hav nething to do with it?

Yes it does.

(x²+7x+12)/(2x²+9x+4)
= (x+4)(x+3)/(2x+1)(x+4)
= (x+3)/(2x+1)

So apparently you don't need to solve the hideous quadratic I posted after all. :biggrin:

So when we get to
(x²+7x+12)/((2x²+9x+4) = e

Unparseable latex formula:

\right

(x+3)/(2x+1) = e
x+3 = (2x+1)e
x+3 = 2ex + e
x(2e-1) = 3-e
x = (3-e)/(2e-1)

Reply 12

OP

I was going to say iv used like 10 pieces of paper trying to solve that quadratic. Got there in the end tho cheers

Reply 13

OP