Author Topic: Recreational Problems (SM level) (Read 102672 times) Tweet Share

ed_saifa · « **Reply #105 on:** March 27, 2008, 06:46:10 pm »

lol. to the thread above mine. That is an ambiguous question =P

Mao · « **Reply #106 on:** March 27, 2008, 07:19:36 pm »

Quote from: DivideBy0 on March 27, 2008, 03:35:13 pm

Prove that if $\log_r p = q$ and $\log_q r = p$ , then $\log_q p = pq$

$\log_r p=q$

$\implies p=r^q$

$\implies log_q p = log_q r^q$

$\implies log_q p=q log_q r$

$\implies log_q p = pq$

QED

ed_saifa · « **Reply #107 on:** March 27, 2008, 07:21:35 pm »

Quote from: Mao on March 27, 2008, 07:19:36 pm

Quote from: DivideBy0 on March 27, 2008, 03:35:13 pm
Prove that if $\log_r p = q$ and $\log_q r = p$ , then $\log_q p = pq$

$\log_r p=q$

$\implies p=r^q$

$\log_q p =pq$

$\implies log_q p = log_q r^q$

$\implies log_q p=q log_q r$

$\implies log_q p = pq$

QED

Wow i like that method of doing that.

Glockmeister · « **Reply #108 on:** March 27, 2008, 08:44:46 pm »

Quote from: Mao on March 27, 2008, 07:19:36 pm

Quote from: DivideBy0 on March 27, 2008, 03:35:13 pm
Prove that if $\log_r p = q$ and $\log_q r = p$ , then $\log_q p = pq$

$\log_r p=q$

$\implies p=r^q$

$\log_q p =pq$

$\implies log_q p = log_q r^q$

$\implies log_q p=q log_q r$

$\implies log_q p = pq$

QED

Damn... makes it seem so easy.

Mao · « **Reply #109 on:** March 27, 2008, 08:52:26 pm »

Quote from: Glockmeister on March 27, 2008, 08:44:46 pm

Damn... makes it seem so easy.

the second one is killing me tho.... =\

Mao · « **Reply #110 on:** March 27, 2008, 09:22:32 pm »

OMG the torture is over thanks to Ahmad :smitten: :smitten: :smitten:
lol here it is

Quote from: DivideBy0 on March 27, 2008, 03:35:13 pm

Suppose a and b are positive numbers for which $\log_9a=\log_{15}b=\log_{25} (a+2b)$

What is the value of b/a?

let $\log_9a=\log_{15}b=\log_{25} (a+2b)=k$

$\implies a=9^k=\left(3^k\right)^2$

$\implies b=15^k=3^k \cdot 5^k$

$\implies a+2b=25^k=\left(5^k\right)^2=\frac{\left(3^k\right)^2}{\left(3^k\right)^2}\cdot \left(5^k\right)^2=\frac{\left( 3^k \cdot 5^k \right)^2}{\left(3^k\right)^2}=\frac{b^2}{a}$

$\implies 1+2\frac{b}{a}=\frac{b^2}{a^2}$

$\therefore \left(\frac{b}{a}\right)^2-2\left( \frac{b}{a} \right) -1=0$

and since $a>0 \ and \ b>0$ , the only solution is

$\frac{b}{a}=1+\sqrt{2}$

*phew* at last....

on a side note divideby0, you're too smart for your own good xD xD I'll be expecting great things from u

/0 · « **Reply #111 on:** March 27, 2008, 09:32:19 pm »

Not really, I couldn't do this problem and that's why I posted it ^^

Glockmeister · « **Reply #112 on:** March 27, 2008, 09:34:01 pm »

Bugger... this is where I got up to (got interrupted by dinner, KFC).

$\log_9a=\log_{15}b=\log_{25} (a+2b)$
$Let C = \log_{9} a$
$\therefore C = \log_{15} b and C = \log_{25} (a+2b)$

Changing into exponential form.

$25^{c} = a +2b$
$9^{c} = a$
$15^{c} = b$
$\therefore 25^{c} = 9^{c} + 2 \times 15^{c}$
$5^{2c} = 3^{2c} + 2 \times 15^{c}$

And this is where I got stuck.

Mao · « **Reply #113 on:** March 27, 2008, 09:35:33 pm »

Glockmeister that is pretty sweet, you thought of setting the C, whereas i had to get Ahmad to hint me....

the key is to try expressing that $25^c$ in terms of a and b

Quote from: DivideBy0 on March 27, 2008, 09:32:19 pm

Not really, I couldn't do this problem and that's why I posted it ^^

^_^ i'll still be expecting great things from you

/0 · « **Reply #114 on:** March 27, 2008, 10:38:06 pm »

Quote from: Mao on March 27, 2008, 07:19:36 pm

Quote from: DivideBy0 on March 27, 2008, 03:35:13 pm
Prove that if $\log_r p = q$ and $\log_q r = p$ , then $\log_q p = pq$

$\log_r p=q$

$\implies p=r^q$

$\implies log_q p = log_q r^q$

$\implies log_q p=q log_q r$

$\implies log_q p = pq$

QED

Alternative:

$\log_q r = \frac{\log_r r}{\log_r q}=\frac{1}{\log_r q}$

So we can multiply together

$\log_r p = q$

and

$\frac{1}{\log_r q}=p$

to get

$\frac{\log_r p}{\log_r q}=pq$

$\log_q p = pq$

Glockmeister · « **Reply #115 on:** March 27, 2008, 11:27:29 pm »

Quote from: DivideBy0 on March 27, 2008, 09:32:19 pm

Not really, I couldn't do this problem and that's why I posted it ^^

I totally agree with Mao's assessment. The fact that you are considering questions like these put you one stead ahead of most of the state. (I'm presuming you're doing Methods 3/4 this year, right?)

Glockmeister · « **Reply #116 on:** March 27, 2008, 11:33:59 pm »

Quote from: Mao on March 27, 2008, 09:35:33 pm

Glockmeister that is pretty sweet, you thought of setting the C, whereas i had to get Ahmad to hint me....

the key is to try expressing that $25^c$ in terms of a and b

Quote from: DivideBy0 on March 27, 2008, 09:32:19 pm
Not really, I couldn't do this problem and that's why I posted it ^^
^_^ i'll still be expecting great things from you

Yeah, I was trying to get it in terms of C, then substitute back into the equations.

Tea.bag · « **Reply #117 on:** March 29, 2008, 07:20:48 pm »

prove this...

If $A+B+C=\pi$ , prove that:

$sinA+sinB+sinC=4 cos\frac{A}{2} cos\frac{B}{2} cos\frac{C}{2}$

Ahmad · « **Reply #118 on:** March 29, 2008, 08:17:52 pm »

$R = 4 \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}$

$= 4 \cos \left(\frac{\pi - B - C}{2}\right) \cos\frac{B}{2} \cos\frac{C}{2}$

$= 4 \cos \frac{B}{2} \cos \frac{C}{2} \left( \sin \frac{B}{2} \cos \frac{C}{2} + \sin \frac{C}{2} \cos \frac{B}{2} \right)$

$= 2 \sin B \cos^2 \frac{C}{2} + 2 \sin C \cos^2 \frac{B}{2}$

Since $\cos^2 \frac{C}{2} = \frac{1}{2} (1 + \cos C)$

$= \sin B + \sin B \cos C + \sin C + \sin C \cos B$

$= \sin (B+C) + \sin B + \sin C$

$= \sin A + \sin B + \sin C = L$

Tea.bag · « **Reply #119 on:** March 29, 2008, 10:47:09 pm »

correct!

another question..

1) if $\begin{bmatrix}4a^2&4a&1\\4b^2&4b&1\\4c^2&4c&1\end{bmatrix}\begin{bmatrix}f(-1)\\f(1)\\f(2)\end{bmatrix}=\begin{bmatrix}3a^2+3a\\3b^2+3b\\3c^2+3c\end{bmatrix}$

$f(x)$ is a quadratic function and its maximum value occurs at a point $V$ . $A$ is a point of

intersection of $y=f(x)$ with x-axis and point $B$ is such that chord $AB$ subtends a right

angle at $V$ . Find the area enclosed by $f(x)$ and chord $AB$ .

2) $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\le1$ and $f'(x)=g(x)$ . If $f^2(0)+g^2(0)=9$ .
Prove that their exists some $c \in (-3,3)$ such that $g(c)$ . $g"(c)<0$

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Author Topic: Recreational Problems (SM level) (Read 102672 times) Tweet Share

ed_saifa

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