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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: dejan91 on February 20, 2009, 05:02:52 pm

Title: Logarithm Help
Post by: dejan91 on February 20, 2009, 05:02:52 pm: 4e^x + 9e^-x = 12

Thanks :)
Title: Re: Logarithm Help
Post by: kamil9876 on February 20, 2009, 05:27:07 pm: multiply every term by e^x. Then you will get some term with e^2x and one with e^x, plus another thats just a constant. let e^x=a and notice that a quadratic forms which you should be able to solve. Once a is solved, logarithms will allow u to find x
Title: Re: Logarithm Help
Post by: dejan91 on February 20, 2009, 05:31:49 pm: Awesome thanks.
Title: Re: Logarithm Help
Post by: dejan91 on March 07, 2009, 11:24:04 am: $T = A\log_{e}x-b$
Solve for $A$ and $b$ , where $T=35.7 when x=0$ , and $T=36.1 when x=10$ .

Help please.
Title: Re: Logarithm Help
Post by: hard on March 07, 2009, 12:49:20 pm: where did the t come from. do you mean when x =0 and x =10?
Title: Re: Logarithm Help
Post by: dejan91 on March 07, 2009, 12:53:06 pm: Oh shit, sorry. The book has T and t, since they represent Temperature and time, but i put in x for t in one part.
Edited it.
Title: Re: Logarithm Help
Post by: Damo17 on March 07, 2009, 07:00:16 pm: Quote from: dejan91 on March 07, 2009, 11:24:04 am
$T = A\log_{e}x-b$
Solve for $A$ and $b$ , where $T=35.7 when x=0$ , and $T=36.1 when x=10$ .

Help please.

Firstly $T=0$ when $x=0$ .
If this is question 11 from Heinemann exercise 2.5:
A patient suffering with hypothermia after being lost in the snow for 2 days is covered in thermal blankets and warmed so that his body temperature can return to normal levels. The change in body temperature $(T^0 C)$ from the initial temperature that occurs during thermal treatment is modelled by the equation $T= A log_e(t-b)$ where t represents the time in minutes. The initial body temp is $35.7^0C$ and it reaches $36.1^0 C$ ten minutes after treatment is initiated.

$0= A log_e(-b)$ ......1
$(36.1-35.7)= A log_e(10-b)$
$0.4=A log_e(10-b)$ ......2

First work out b:
$0= A log_e(-b)$
$e^0=-b$
$b=-1$

Work out A now:
$0.4=A log_e(10-(-1))$
$A= 0.4/log_e (11)$
$= 0.169$

They are the right answers for this question from heinemann.
Title: Re: Logarithm Help
Post by: dejan91 on March 08, 2009, 11:15:55 am: Yup that was the Heinemann question I just cbf'd writing the whole thing out. Thanks for that by the way.
Title: Re: Logarithm Help
Post by: dejan91 on April 21, 2009, 09:14:34 pm: Help?

Given that $f(x) = e^x sin(a)$ where $0 \le a \le \frac{\pi}{2}$ , find the value of $a$ such that $f'(x) = 0.5$ .
Title: Re: Logarithm Help
Post by: Mao on April 21, 2009, 09:24:27 pm: $f'(x) = 0.5$ ...? x is not specified?
Title: Re: Logarithm Help
Post by: TrueTears on April 21, 2009, 09:24:36 pm: $f'(x) = e^xsin(a)$ (treating sin(a) as a constant because we are differentiating with respect to x)

$e^xsin(a) = 0.5$

$sin(a) = \frac{1}{2e^x}$

$a = sin^{-1}(\frac{1}{2e^x})$
Title: Re: Logarithm Help
Post by: dejan91 on April 21, 2009, 10:14:38 pm: Ah damn sorry Mao, meant to be $f(0) = 0.5$ not $f(x) = 0.5$ . But now I see, so when you substitue x = 0, a = 0.524 (which is what the book had).

The part that confused me was that I didn't know to keep $sin(a)$ as a constant. I get it now thanks TT :)
Title: Re: Logarithm Help
Post by: dejan91 on June 25, 2009, 12:48:45 am: Simplify: $3x(1+log_{e}x^{3} + 2log_{e}x)$ . I get to a point where I'm just not sure what to do, or even if it's right.
Title: Re: Logarithm Help
Post by: Flaming_Arrow on June 25, 2009, 08:23:12 am: $3x(1+log_{e}x^{3} + 2log_{e}x)$

$3x(1+log_{e}x^{3} + log_{e}x^2)$

$3x(1+log_{e}({x^{3}*x^2))$

$3x(1+log_{e}({x^5))$

$3x(1+5 log_{e}{x})$

im not sure if u can simplify it further
Title: Re: Logarithm Help
Post by: TrueTears on June 25, 2009, 04:44:01 pm: Quote from: Flaming_Arrow on June 25, 2009, 08:23:12 am
$3x(1+log_{e}x^{3} + 2log_{e}x)$

$3x(1+log_{e}x^{3} + log_{e}x^2)$

$3x(1+log_{e}({x^{3}*x^2))$

$3x(1+log_{e}({x^5))$

$3x(1+5 log_{e}{x})$

im not sure if u can simplify it further
Further simplify:

$1 = log_e(e)$

Hence $3x(log_e(e) + log_e(x^5))$

$3x(log_e(ex^5))$

But I'm sure FA's final answer is perfectly acceptable.
Title: Re: Logarithm Help
Post by: dejan91 on June 25, 2009, 10:27:52 pm: Ok thanks guys :)

What about $3x(1+log_{e}x^{3} - 2log_{e}x)$ ?
Title: Re: Logarithm Help
Post by: TrueTears on June 25, 2009, 10:30:58 pm: $3x(1+ log_ex^3 - log_ex^2)$

$3x(1 + log_e(\frac{x^3}{x^2}))$

$3x(1 + log_e(x))$

You could do the 1 step further like I did before, but I don't think it's necessary.
Title: Re: Logarithm Help
Post by: dejan91 on June 25, 2009, 10:38:06 pm: Well, I got that far. But the problem was that it was a multiple choice question. Options were:
$3x + log_e(x)$
$3xlog_e(x)$

...And that's all I remember. It was on my sac I had and it's been bugging me ever since.
Title: Re: Logarithm Help
Post by: TrueTears on June 25, 2009, 10:39:48 pm: $3x(log_ee + log_ex)$

$3x(log_eex)$ which seems like one of the answers you remember.

or $3x + 3xlog_ex$ which seems like another one of the answers you remember.

It's either one of those .
Title: Re: Logarithm Help
Post by: dejan91 on June 25, 2009, 10:43:15 pm: Yup, it was definately the first one...There goes two marks lol oh well. Thanks :)