Ok, finally got to grips with the advanced algebra part of my coursework.

Or so i thought. Now i reckon this is easy and my brain has reached saturation point but im really struggling with this one.

Oh and realman if u read this. Cheers for the quadratic equation advice. Made everything so simple 🙂

Anyhoo i have hopefully attached a pic of the offending question. Please help 🙂

Is that the whole equation? I don’t really see what is there to solve and unless I’ve forgotten how to use my calculator (Which is very possible), the rhs is not 16…

Unless e is an unknown and not what I think of as e…

I think ‘e’ is a variable in this case like ‘x’ would be, rather than the normally associated constant.

(Otherwise, the correct answer would be ‘false’).

Yep thats the whole equation, e relates to growth and is 2.7182.

Now my assumption was to transform and solve it for t.

a quick in the head says

t = -2 * ln(1/3)

e^(-t/2)=1/3

I can’t remember how to take it further…

Oh I though the t was a 1… That makes more sense.

Quick in the head is the same as klunk but with a minus sign in front. I was always prone to getting minus signs wrong though

Edit. Didn’t see the minus on klunks answer. That’s what I get as well.

Yea it has pickled my head for now. Just gotta sleep on it and see if that does me any good. i spent 3 bloody days figuring out linear equations with 3 variables so this should be a doddle 🙂

klunk has it I think.

Just remembering old math:

16=24(1-e^(-t/2))

16 = 24 – 24e^(-t/2)

-8 = -24 e^(-t/2)

multiply by -1 both sides:

8 = 24 e^(-t/2)

1/3 = e^(-t/2)

take logs:

ln(1/3) = (-t/2)*ln(e)
-2ln(1/3)/ln(e)=t

But ln(e) = 1
-2 ln(1/3) = t

Which makes t=+2.19722457733624 (since ln(x) where x<1 is negative).

Hi Emsz, if you’re about!