Author Topic: VCE Methods Question Thread! (Read 5782252 times) Tweet Share

Conic · « **Reply #1875 on:** April 27, 2013, 03:08:25 pm »

Simplify it:

$10=ae^{-2ln(5)}$

Using log laws:

$k*ln(x) = ln(x^k)$

$-2ln(5)=ln(5^{-2})=ln(\frac{1}{5^2})=ln(\frac{1}{25})$

So now you can write it as:

$10=e^{ln(\frac{1}{25})$

Using the definition of a logarithm:

$ln(n)=x \Rightarrow e^x=n \therefore e^{ln(x)}=x$

$10=a(\frac{1}{25})$

$10*25=a$

$250=a$

shadows · « **Reply #1876 on:** April 27, 2013, 03:20:41 pm »

Quote from: Conic on April 27, 2013, 03:08:25 pm

Simplify it:

$10=ae^{-2ln(5)}$

Using log laws:

$k*ln(x) = ln(x^k)$

$-2ln(5)=ln(5^{-2})=ln(\frac{1}{5^2})=ln(\frac{1}{25})$

So now you can write it as:

$10=e^{ln(\frac{1}{25})$
Sorry I don't know how you get that?

Using the definition of a logarithm:

$ln(n)=x \Rightarrow e^x=n \therefore e^{ln(x)}=x$

$10=a(\frac{1}{25})$

$10*25=a$

$250=a$

Alwin · « **Reply #1877 on:** April 27, 2013, 03:28:21 pm »

Quote from: shadows on April 27, 2013, 03:20:41 pm

Quote from: Conic on April 27, 2013, 03:08:25 pm
$-2ln(5)=ln(5^{-2})=ln(\frac{1}{5^2})=ln(\frac{1}{25})$

So now you can write it as:

$10=e^{ln(\frac{1}{25})$
Sorry I don't know how you get that?

All conic did was simply,

$10=ae^{-6(\frac{1}{3}loge(5))$
$10=ae^{-2ln(5)}$
$10=ae^{ln(5^{-2})}$
$10=ae^{ln(\frac{1}{25})$
$10=a\cdot \frac{1}{25}$

$\therefore a=250$

Make more sense?

shadows · « **Reply #1878 on:** April 27, 2013, 03:29:44 pm »

$10=ae^{loge\frac{1}{25}}$

is equal to $10=a(\frac{1}{25})?$

loge(e) cancels out and $\frac{1}{25}$ comes down right?

Zealous · « **Reply #1879 on:** April 27, 2013, 04:28:11 pm »

Quote from: shadows on April 27, 2013, 03:29:44 pm

$10=ae^{loge\frac{1}{25}}$

is equal to $10=a(\frac{1}{25})?$

loge(e) cancels out and $\frac{1}{25}$ comes down right?

Correcto.
$e^{loge({x})} = x$

..so you can factor out $a$ and cancel out $e$ and $loge$

shadows · « **Reply #1880 on:** April 27, 2013, 05:09:38 pm »

Thank you!

can someone explain how

$\frac {loge^{32}}{loge^2}$ is equal to 5

I realised this works with whatever base you change, and the bottom (2) powered by 5 (answer) is equal to top (32)

LOL i think I might be overthinking this but I am absolute terrible at logs.... and I can only remember these things if I understand it. I will forget if I blindlessly rote learn it.

I'm kind of stressed because my SACS are coming up D:

Anyone have any useful log/ exp laws (another than the standard ones) ..if that makes sense. Anyone have any tips? Or just do more questions..?

b^3 · « **Reply #1881 on:** April 27, 2013, 05:21:01 pm »

The rule for the change of base of a log is $\begin{alignedat}{1}\log_{a}x & =\frac{\log_{b}x}{\log_{b}x}\end{alignedat}$ . Lets say we start off with $y=log_{a}(x)$ , which is the same as saying $a^{y}=x$ .
Take the log with base $b$ of both sides, we obtain
$\begin{alignedat}{1}a^{y} & =x \\ \log_{b}\left(a^{y}\right) & =\log_{b}\left(x\right) \\ y\log_{b}\left(a\right) & =\log_{b}\left(x\right) \\ y & =\frac{\log_{b}\left(x\right)}{\log_{b}\left(a\right)} \\ \implies\log_{a}\left(x\right) & =\frac{\log_{b}\left(x\right)}{\log_{b}\left(a\right)} \end{alignedat} $
So here we have
$\begin{alignedat}{1}\frac{\log_{e}32}{\log_{e}2} & =\log_{2}32 \\ & =\log_{2}2^{5} \\ & =5\log_{2}2 \\ & =5 \end{alignedat} $

Quote from: shadows on April 27, 2013, 05:09:38 pm

Anyone have any useful log/ exp laws (another than the standard ones) ..if that makes sense. Anyone have any tips? Or just do more questions..?

Depend by what you mean by the standard ones?
I think you might be thinking of things like $e^{\log_{e}(x)}$ ?
$\begin{alignedat}{1}\text{Let }a & =e^{\log_{e}\left(x\right)},\: x>0 \\ \log_{e}\left(a\right) & =\log_{e}\left(e^{\log_{e}\left(x\right)}\right) \\ \log_{e}\left(a\right) & =\log_{e}\left(x\right)\log_{e}\left(e\right) \\ \log_{e}\left(a\right) & =\log_{e}\left(x\right) \\ \implies x & =a,\: x>0 \\ \implies e^{\log_{e}(x)} & =x,\: x>0 \end{alignedat} $

shadows · « **Reply #1882 on:** April 27, 2013, 05:43:33 pm »

Thanks so much!

It understand it now!

shadows · « **Reply #1883 on:** April 27, 2013, 07:11:37 pm »

lol sorry another question xD

$3^{2x}-3^{x+2}+8=0$

How do i solve for x? It says I need to solve by change base?

thank you

darklight · « **Reply #1884 on:** April 27, 2013, 07:20:10 pm »

Quote from: shadows on April 27, 2013, 07:11:37 pm

lol sorry another question xD

$3^{2x}-3^{x+2}+8=0$

How do i solve for x? It says I need to solve by change base?

thank you

I wouldn't change base for this one. What I would do is split $3^{x+2}$ into $3^x * 3^2$ .

Then, let $3^x=a$ .

Can you go from here?

Conic · « **Reply #1885 on:** April 27, 2013, 07:25:39 pm »

Quote from: shadows on April 27, 2013, 07:11:37 pm

lol sorry another question xD

$3^{2x}-3^{x+2}+8=0$

How do i solve for x? It says I need to solve by change base?

thank you

$-3^{x+2}=-3^x*3^2=-9*3^x$

So you can write the question as:

$3^{2x}-9*3^x+8$

let $3^x=u$

$u^2-9u+8=0$

$(u-1)(u-8)=0$

so u=1 or u=8

$1=3^x \Rightarrow x=0$ or $8=3^x \Rightarrow x=log_3 (8)$

$x=0, x=log_3 (8)$

You can also write $log_3 (8)$ as $3log_3 (2)$

shadows · « **Reply #1886 on:** April 27, 2013, 08:56:31 pm »

Quote from: Conic on April 27, 2013, 07:25:39 pm

$-3^{x+2}=-3^x*3^2=-9*3^x$

So you can write the question as:

$3^{2x}-9*3^x+8$

let $3^x=u$

$u^2-9u+8=0$

$(u-1)(u-8)=0$

so u=1 or u=8

$1=3^x \Rightarrow x=0$ or $8=3^x \Rightarrow x=log_3 (8)$

$x=0, x=log_3 (8)$

You can also write $log_3 (8)$ as $3log_3 (2)$

Quote from: darklight on April 27, 2013, 07:20:10 pm

I wouldn't change base for this one. What I would do is split $3^{x+2}$ into $3^x * 3^2$ .

Then, let $3^x=a$ .

Can you go from here?

Thanks guys! Hmm weird question.. so you can't solve it by changing base?

Smiley_ · « **Reply #1887 on:** April 28, 2013, 12:00:49 pm »

The height of the arch is 9 metres (OZ = 9 m). The width of the arch is 20 metres (AB=20m).Theequationofthecurveisoftheform y =ax2 +b,takingaxesas shown.
a Find values of a and b.
b Amanofheight1.8mstandsatC(OC=7m).HowfarabovehisheadisthepointE
on the arch? (That is, find the distance DE.)
c A horizontal bar FG is placed across the arch as shown. The height, OH, of the bar above the ground is 6.3 m. Find the length of the bar.

So got that b = 9 but I don't know what I do to get the other point

Smiley_ · « **Reply #1888 on:** April 28, 2013, 12:25:42 pm »

Ok so I figured out a but I'm still confused on b it's chapter 8 question 1'esstentials

Conic · « **Reply #1889 on:** April 28, 2013, 12:36:15 pm »

b:
The height of the bridge 7m from the ground is the y value of the function at 7.

$y=\frac{-9}{100}7^2+9$

$y=9-\frac{441}{100}$

$y=9-4.41$

$y=4.59$

The heigt of the arch above his head is the difference between his height and the height of the arch ie 4.59-1.8, which is 2.79m.

c:
You know that the height is 6.3m, so that means the y value is 6.3. You need to solve for x:

$6.3=\frac{-9}{100}x^2+9$

$2.7=\frac{9}{100}x^2$

$x^2=30$

$x=\pm \sqrt{30}$

The length of the bar is the distance between the 2 points, ie $\sqrt{30}--\sqrt{30}=2\sqrt{30}$ , which is approximately 10.95m.

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