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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Halil on March 16, 2011, 05:40:41 pm

Title: 2 questions
Post by: Halil on March 16, 2011, 05:40:41 pm: We had a test today and I'm sure i got the following two wrong.
Can you please show the right solutions?

1st is 3e^6-x and e^x-2 both intersect at 4+0.5ln(3)
Show.

2nd is 3^x-1=2^x+a. Solve for x when a is a subset of R+
Title: Re: 2 questions
Post by: onur369 on March 16, 2011, 06:16:46 pm: For the second one I did log3(3)^x-1=log2(2^3)^x+a

log3(3) and log2(2) both equal one, move the squared values down.
it will be
x-1=3x+a
2x=-1-a
x=(-1-a)/2

I dont know if its correct though -.-
Title: Re: 2 questions
Post by: luffy on March 16, 2011, 07:21:55 pm: Below is how I would do your questions:
1) $3e^{6-x} = e^{x-2}$
Divide both sides by $e^{x-2}$
$\frac{3e^{6-x}}{e^{x-2}} = 1$
$3e^{(6-x)-(x-2)} = 1$
$e^{8-2x} = \frac{1}{3}$
$8-2x = log_{e}(\frac{1}{3})$
$8-2x = -log_{e}(3)$
$2x = 8 + log_{e}(3}$
$x = 4+ 0.5log_{e}{3}$
-> As required.

2) $3^{x-1} = 2^{x + a}$
In methods, you are allowed to "log both sides if you use the SAME base". (i.e. what onur369 did... doesn't really make sense to me)
For example, 3 = x means log_{3}{3} = log_{3}(x) -> that is true.

$\therefore log_{3}(3^{x-1}) = log_{3}(2^{x+a})$
$(x-1)log_{3}(3) = (x+a)log_{3}(2)$
As $log_{3}(3) = 1$
$x - 1 = (x + a)log_{3}(2)$
$x -1=xlog_{3}(2) + alog_{3}(2)$
$x - xlog_{3}(2) = alog_{3}(2) + 1$
$x(1 - log_{3}(2)) = alog_{3}(2) + 1$
$x = \frac{alog_{3}(2) + 1}{1 - log_{3}(2)}$

I might have made a slight error somewhere in my working of the second one. Can someone please confirm the accuracy of my answer? I'm not sure if you are meant to simplify it any further, but lets see how it goes:
$x = \frac{log_{3}(3\times2^a)}{log_{3}(1.5)})$
$x = {log_{1.5}(3\times2^a)$
If you want it in terms of "e", you could always just say:
$x = \frac{log_{e}(3\times2^a)}{log_{e}(1.5)})$

Hope I helped.
Title: Re: 2 questions
Post by: Halil on March 16, 2011, 07:57:07 pm: :( :( :(
I knew the solution to the first one but I couldnt remember it at the time.
Thanks for the help luffy :)

However, there was another question similar to the first one which i substituted the value of x in to both of the equations and i got the same answer.
I tried to do the same for this one. But it did not work.
Can you show that way too please?
Title: Re: 2 questions
Post by: luffy on March 17, 2011, 09:52:34 pm: Quote from: Halil on March 16, 2011, 07:57:07 pm
:( :( :(
I knew the solution to the first one but I couldnt remember it at the time.
Thanks for the help luffy :)

However, there was another question similar to the first one which i substituted the value of x in to both of the equations and i got the same answer.
I tried to do the same for this one. But it did not work.
Can you show that way too please?

I'm not quite sure what you meant. Could you go into a bit more detail? I can't really think of another method of approaching this question (using knowledge obtained from methods).