VCE Methods Question Thread!

[keltingmeith]

Alternatively to theshunpo, when you see fractions where the bigger power is on top of the smaller one, then you can do long division to simplify it out. Do this when you see if it applies since it would easily solve this question for you I guess.

Basically that equation would simply out to 3x-3 and if you take the derivative and sub x=2, you'd get 3 as your answer still

On this:

[theshunpo]

IndefatigableLover and EulerFan101, your method for working this question out are so much more efficient than mine . I guess i just did the first thing that came to mind haha

[keltingmeith]

IndefatigableLover and EulerFan101, your method for working this question out are so much more efficient than mine . I guess i just did the first thing that came to mind haha

Don't worry - if indefatigable hadn't pointed out the long division method, I probably would've missed that and done the quotient rule as you did.

[Damo23]

Hey how do u guys work this out
3*2^2x-2^x=0 solve for x

And also

2log10(x-3)=10 solve for x

Thanks

[brightsky]

For the first one:

Let A = 2^x. The equation becomes 3A^2 - A = 0, which implies A (3A - 1) = 0. Hence, A = 0 or A = 1/3, which implies 2^x = 0 or 2^x = 1/3. Since 2^x > 0 for all x, 2^x cannot be 0. So 2^x = 1/3, which implies x = log_2(1/3) = -log_2(3).

For the second one:

log_10(x-3) = 5 implies 10^5 = x - 3, so x = 10^5 + 3, which is an exceedingly large number...

EDIT: Typo rectified.

[Damo23]

For the first one:

Let A = 2^x. The equation becomes 3A^2 - A = 0, which implies A (3A - 1) = 0. Hence, A = 0 or A = 1/3, which implies 2^x = 0 or 2^x = 1/3. Since 2^x > 0 for all x, 2^x cannot be 0. So 2^x = 1/3, which implies x = log_2(1/3) = -log_2(3).

For the second one:

log_10(x-3) = 5 implies 10^5 = x - 3, so x = 10^5 - 3, which is an exceedingly large number...

oh thanks and for the second one did you get =5 by dividing by the 2?

[theshunpo]

oh thanks and for the second one did you get =5 by dividing by the 2?

Yeah he did

[Damo23]

Yeah he did

sorry to bother you but why do u divide the =10 by 2, because other times I've squared the one in the  brackets like (x-3) is this case.

Sorry I didn't really understand log to well

[keltingmeith]

sorry to bother you but why do u divide the =10 by 2, because other times I've squared the one in the  brackets like (x-3) is this case.

Sorry I didn't really understand log to well

This is all about re-arranging equations - the fact that there's a logarithm doesn't make the exercise any different. Don't worry, though - you're not the only to be scared by the introduction of new functions and be unsure about what's going on - I have yet to meet a methods student this HASN'T happened to!

Instead of thinking, "I'm solving an equation with logs!", instead I want you to pretend that the log is just another number.

So, you had 2log_10(x-3) = 10 - pretend that "log_10(x-3)" is a variable (say, u). This means we have 2u=10. Now, if you wanted to solve for u, any other time you'd divide by the 2 - so, let's do that, giving us u=5.

At this point, you'd remember that u is a logarithm you need to solve - and so you'd employ the techniques used before. In fact, a simple guide to solving logarithms:

-Solve for the logarithm just like you would any other variable.
-Once you have that logarithm by itself, raise both sides to the power of the logarithm's base to remove the logarithm.

How you get that logarithm by itself is irrelevant - in fact, here's another method using log rules (which is what I think you were getting at?):



At this point, we'd need to check both answers in the original equation - just in case we accidentally added some extra answers. The first answer works fine, but the second is a negative number - and so won't work inside of the log, which means we need to disclude it from our solution. So, as before, we get

[Damo23]

This is all about re-arranging equations - the fact that there's a logarithm doesn't make the exercise any different. Don't worry, though - you're not the only to be scared by the introduction of new functions and be unsure about what's going on - I have yet to meet a methods student this HASN'T happened to!

Instead of thinking, "I'm solving an equation with logs!", instead I want you to pretend that the log is just another number.

So, you had 2log_10(x-3) = 10 - pretend that "log_10(x-3)" is a variable (say, u). This means we have 2u=10. Now, if you wanted to solve for u, any other time you'd divide by the 2 - so, let's do that, giving us u=5.

At this point, you'd remember that u is a logarithm you need to solve - and so you'd employ the techniques used before. In fact, a simple guide to solving logarithms:

-Solve for the logarithm just like you would any other variable.
-Once you have that logarithm by itself, raise both sides to the power of the logarithm's base to remove the logarithm.

How you get that logarithm by itself is irrelevant - in fact, here's another method using log rules (which is what I think you were getting at?):



At this point, we'd need to check both answers in the original equation - just in case we accidentally added some extra answers. The first answer works fine, but the second is a negative number - and so won't work inside of the log, which means we need to disclude it from our solution. So, as before, we get

thanks so much Eulernfan101 you really helped clear my mind and understand it, thanks heaps!!!   

[Deshouka]

Hi.
Can anyone help me with question 10C in methods 2013 exam 1?
I've looked at the worked solutions but I don't quite understand it.
Thanks

[theshunpo]

Hi.
Can anyone help me with question 10C in methods 2013 exam 1?
I've looked at the worked solutions but I don't quite understand it.
Thanks

The two ways I see that you could do that question is:

1) Work out the equation of ST then find the area bounded by taking the definite integral of ST-f(x) between the points of intersection

OR (my preferred way)

2) Find the area of the trapezium and subtract it from the area under f(x) (you'll first have to find the x-coordinate of where f(x)=1/2) then integrate between 0 and that point.

[anna.xo]

Really general question - but how do I clear the memory of my cas ? TIA

[keltingmeith]

Really general question - but how do I clear the memory of my cas ? TIA

Pretty sure more TI-nspires have a reset button on the back (the kind you need a pin for). Not sure about the ClassPad...

[knightrider]

Is there a specific order set by VCAA to do the methods course .

Because my school seems to be doing the methods course in different order chapters not like chapter 1 - 12

Just wanted to know