Author Topic: VCE Methods Question Thread! (Read 6006085 times) Tweet Share

psyxwar · « **Reply #8490 on:** February 07, 2015, 09:27:21 pm »

Quote from: warya on February 07, 2015, 08:57:56 pm

Can someone help me with this question

2^(x-1)=3^(x+a ) for a, where a is an element of R

$2^{(x-1)}=3^{(x+a)}$
$\implies \log_{3}(2^{(x-1)})=\log_{3}(3^{(x+a)})$
$\implies x+a=(x-1)\log_{3}(2)$
$\implies a=x\log_{3}(2)-\log_{3}(2)-x$
$\implies a=(\log_{3}(2)-1)x-\log_{3}(2)$

Quote

and this one

log7(x)=log4(p) for x

$\log_{7}(x)=\log_{4}(p)$
$\implies 7^{\log_{7}(x)}=7^{\log_{4}(p)}$
$\implies x=7^{\log_{4}(p)}{$

edit: ninja'd

warya · « **Reply #8491 on:** February 07, 2015, 09:34:13 pm »

I just realised for question 1 I meant solve for x

Thanks for the second one!

warya · « **Reply #8492 on:** February 07, 2015, 09:38:49 pm »

Sorry guys the textbooks answer for the second one, log7(x)=log4(p)
is x=7^(log e (p)/log e (4))

I did the way you guys showed me as well but I was confused by the answer in the textbook?

psyxwar · « **Reply #8493 on:** February 07, 2015, 09:39:58 pm »

Quote from: warya on February 07, 2015, 08:57:56 pm

Can someone help me with this question

2^(x-1)=3^(x+a ) for a, where a is an element of R

$\implies (x-1)\ln(2)=(x+a)\ln(3)$
$\implies \frac{x-1}{x+a}=\frac{\ln(3)}{\ln(2)}$
$\implies \frac{(x+a)-(a+1)}{x+a}=\frac{\ln(3)}{\ln(2)}$
$\implies 1-\frac{1+a}{x+a}=\frac{\ln(3)}{\ln(2)}$
$\implies \frac{1+a}{x+a}=1-\frac{\ln(3)}{\ln(2)}=\frac{\ln(2)-\ln(3)}{\ln(2)}$
$\implies \frac{x+a}{1+a}=\frac{\ln(2)}{\ln(2)-\ln(3)}$
$\implies x+a = \frac{(1+a)\ln(2)}{\ln(2)-\ln(3)}$
$\implies x=\frac{(1+a)\ln(2)}{\ln(2)-\ln(3)}-a$

psyxwar · « **Reply #8494 on:** February 07, 2015, 09:42:21 pm »

Quote from: warya on February 07, 2015, 09:38:49 pm

Sorry guys the textbooks answer for the second one, log7(x)=log4(p)
is x=7^(log e (p)/log e (4))

I did the way you guys showed me as well but I was confused by the answer in the textbook?

They changed it to a log base e using the change of base formula (not sure why). The answers are equivalent.

$\log_{4}(p)=\frac{\log_{e}(p)}{\log_{e}(4)}$

warya · « **Reply #8495 on:** February 07, 2015, 09:46:24 pm »

Oh I see thanks

and for the one you just posted, is it the same as the answer in the textbook

x=(aln(3)+ln(2))/ln(2/3)

keltingmeith · « **Reply #8496 on:** February 07, 2015, 09:50:07 pm »

Quote from: psyxwar on February 07, 2015, 09:42:21 pm

They changed it to a log base e using the change of base formula (not sure why). The answers are equivalent.

$\log_{4}(p)=\frac{\log_{e}(p)}{\log_{e}(4)}$

Just for why:

Calculators like to compute things to the base e (because of Taylor series, if you're curious). As such, if they used a calculator to find the answer for them, it'll spit something out in base e. Hell, if they're just an "old" mathematician, they're probably just used to doing it because that's how their old log tables gave them answers... (but those might've been in base 10, as well)

psyxwar · « **Reply #8497 on:** February 07, 2015, 09:51:14 pm »

Quote from: warya on February 07, 2015, 09:46:24 pm

Oh I see thanks

and for the one you just posted, is it the same as the answer in the textbook

x=(aln(3)+ln(2))/ln(2/3)

$x=\frac{(1+a)\ln(2)}{\ln(2)-\ln(3)}-a$
$\implies x= \frac{\ln(2)+a\ln(2)}{\ln(\frac{2}{3})}-\frac{\ln(\frac{2}{3})a}{\ln(\frac{2}{3})}$
$\implies x=\frac{a(\ln(2)-\ln(\frac{2}{3}))+\ln(2)}{\ln(\frac{2}{3})}$
$\implies x= \frac{a\ln(3)+\ln(2)}{\ln(\frac{2}{3})}$

So yeah they're equivalent

warya · « **Reply #8498 on:** February 07, 2015, 09:53:00 pm »

Thanks heaps guys

IndefatigableLover · « **Reply #8499 on:** February 07, 2015, 10:07:45 pm »

Quote from: lisax3 on February 07, 2015, 07:40:11 pm

Can someone please help me with this question
Find the inverse of the function f: R\{3/2} R, f(x)=(x + 1) / (2x – 3). State the domain and range of the inverse function.

$f(x)=\frac{x+1}{2x-3}$
Use long division to turn it into something we can use!

$\therefore \frac{x+1}{2x-3}=\frac{1}{2}+\frac{5}{2(2x-3)}$

Now we'll let f(x) = y and then we interchange the two and transpose to find 'y'

$\therefore x = \frac{1}{2}+\frac{5}{2(2y-3)}$

$x-\frac{1}{2}=\frac{5}{2(2y-3)}$

$\frac{1}{x-\frac{1}{2}}=\frac{2(2y-3)}{5}$

$\frac{2}{2x-1}=\frac{2(2y-3)}{5}$

$\frac{10}{2x-1}=4y-6$

$4y=\frac{10}{2x-1}+6$

$4y=\frac{12x+4}{2x-1}$

$y=\frac{4(3x+1)}{2x-1} \times \frac{1}{4}$

$\therefore f^{-1}(x)=\frac{3x+1}{2x-1}$

Now domain is that 'x' cannot equal 1/2 (as seen by 2x-1).
Utilising the domain for f(x) which we know was all real numbers except 3/2, this is also our range (that is 'y' can take all real numbers except 3/2).

x: R\{1/2}
y: R\{3/2}

kinslayer · « **Reply #8500 on:** February 07, 2015, 11:15:13 pm »

Quote from: IndefatigableLover on February 07, 2015, 10:07:45 pm

$f(x)=\frac{x+1}{2x-3}$
Use long division to turn it into something we can use!

$\therefore \frac{x+1}{2x-3}=\frac{1}{2}+\frac{5}{2(2x-3)}$

And if you would prefer not to use long division:

$\frac{x+1}{2x-3} = \frac{1}{2} \left[ \frac{(x - \frac{3}{2}) + \frac{5}{2}}{x-\frac{3}{2}}\right] = \frac{1}{2} + \frac{5/2}{2x - 3}$

warya · « **Reply #8501 on:** February 08, 2015, 12:01:16 am »

How does x-1/x+a become (x+a-a+1)/x+a

kinslayer · « **Reply #8502 on:** February 08, 2015, 12:12:40 am »

Quote from: warya on February 08, 2015, 12:01:16 am

How does x-1/x+a become (x+a-a+1)/x+a

It doesn't. Also please use brackets when writing fractions. What you've written there is $x - \frac{1}{x} + a$ .

knightrider · « **Reply #8503 on:** February 08, 2015, 12:26:11 am »

For this question the infinity is meant to be in round brackets right instead of in square brackets?

What transformations are required to obtain $\sqrt{-x-3}$ from $\sqrt{x-4}.$ ?

1.i got translation of 1 unit in the negative direction of the x-axis.
2.reflection in the y-axis.

but the book got 1.Reflection in the y-axis, 2.translated +1
parallel to the x-axis.

Who would be right?

pi · « **Reply #8504 on:** February 08, 2015, 12:29:09 am »

Quote from: knightrider on February 08, 2015, 12:26:11 am

For this question the infinity is meant to be in round brackets right instead of in square brackets?

Yeah it should always be in round brackets.

I'll leave transformations for someone who can remember maths but I'm fairly sure the book is right just from quickly graphing it in my head!

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