Bazza's 3/4 Question Thread

[dc302]

As always thanks for the help

Why cant a log to the base of a negative work?

If you have doesnt x = 2? why cant you rearrange it and solve using the log function in the cas?

Oh and as for solving it on your calculator, you can but you will get complex solutions, as a result of:

log(-2)4 = log[4]/log[-2] => complex

[WhoTookMyUsername]

Thanks guys.

All the answers written in the text are of the form log_e(x).

Is ln(x) still fine?

[pi]

Is ln(x) still fine?

Yeah, that's fine

Having said that, "lg(x)" for base 10 log(x) and "lb(x)" for base 2 log(x) are NOT fine

[WhoTookMyUsername]

How do i solve this?
I've looked at TT's advice (on inequalities) but i don't really understand it

≥


thanks

understand now xD

[WhoTookMyUsername]

How do i diff this by hand D: keep getting wrong answer according to the sac we just had, not sure if the sac was wrong i i was, chucked it in cas, didnt want to mess around with the answer i case of incorrect working again

Thanks

[pi]

What exactly do you want to do to that?


edit: didn't see your edit, you could use the chain rule (u=(x-1)/(x+1)) and then the quotient rule to find d/du imo.

[paulsterio]

it's root(x-1)/root(x+1) then use the quotient rule

[WhoTookMyUsername]

Yeah i got an (x+1)^(3/2) down the bottom, in the sac it basically told us the answer was (x+1)^2 down the bottom, not sure where i went wrong

[BlueSky_3]

Yeah I got (x+1)^2 in the bottom aswell....

[WhoTookMyUsername]

Actually cas seems to agree with me... Ish... Except it appears to be missing a negative D:

[b^3]

This is what I'm getting, same as what the CAS calc is saying


I'm not sure why it would suggest that you would get (x+1)^2 down the bottom unless you look at it before you simplify it down a bit more (i.e. the second line)

[Somye]

At the first step, rationalise the denominator, and work from there

[WhoTookMyUsername]

something dodgy is going on

on the methods sac we had it implied that that was the answer except 2 replaces 3/2 and -1 replaces 1 (up the top)

and i had that answer except for a negative there

oh wait
um
isn't product rule u/v = (vu' - uv')/v^2
(you have u to the power of 2?)

[Somye]

ps. the answer in the sac was

[paulsterio]

Easiest way to check if the answer in the SAC is right is by doing the following, differentiate it on the cas, now substitute in a really random and weird number, now, I'd say pick 12.5, just a random number that will produce real answers, now set it to Decimal and get the answer. Now substitute 12.5 into the answer on the SAC and see if you get the same answer. If it's the same, it's likely to be equivalent, just because 12.5 is such an obscure number and is not critical.