VCE Methods Question Thread!

[jazzycab]

In reality, there's little point putting it on exam 2 if you can easily check your answer.

The ones that have appeared on exam 2 in the past have typically been multiple choice questions where the integrals have been given in terms of an unspecified function \(f\left(x\right)\)

[snowisawesome]

x = ((log_e(2n)-m)/(-k)

x = ((m-log_e(2n))/(k)

Are both of the above the same?

To sketch y = log_e(x+4) without a cas,
the y intercept would be log_e(4)
Is there a way to work out log_e(4) without a cas?

[Sine]

x = ((log_e(2n)-m)/(-k)

x = ((m-log_e(2n))/(k)

Are both of the above the same?

To sketch y = log_e(x+4) without a cas,
the y intercept would be log_e(4)
Is there a way to work out log_e(4) without a cas?

to check whether two expressions are the same type both into your cas in the form A = B and press enter if it says true it means the equation is valid and both are the same.

You don't need to evaluate ln(4) since the exact form is what is preffered (you lose marks for having non-exact decimals) when writing out intercepts.

[snowisawesome]

to check whether two expressions are the same type both into your cas in the form A = B and press enter if it says true it means the equation is valid and both are the same.

You don't need to evaluate ln(4) since the exact form is what is preffered (you lose marks for having non-exact decimals) when writing out intercepts.

But don't you still need to know the exact value of  ln(4) to know where it is located on a graph?

[Sine]

But don't you still need to know the exact value of  ln(4) to know where it is located on a graph?

you don't really need to know ln(4) = 1.38 you should just be able to determine that it is between 1 and 2 because e^1 = 2.718 and e^2 > 4

[snowisawesome]

you don't really need to know ln(4) = 1.38 you should just be able to determine that it is between 1 and 2 because e^1 = 2.718 and e^2 > 4

But isn't log_e different from e^1

[Bell9565]

But isn't log_e different from e^1

For the sake of graphing you simply need to know where it is approximately, and if e^x is is equal to 4, and e^1 is 2.7 and e^2 is around 7ish then it can be concluded that it would be between 1&2

[snowisawesome]

For the sake of graphing you simply need to know where it is approximately, and if e^x is is equal to 4, and e^1 is 2.7 and e^2 is around 7ish then it can be concluded that it would be between 1&2

Thanks Bell9565 and Sine
Had another question
solve 2^x < 0.3
worked solutions from textbook
log₂(0.3) < x
-1.737 > x
x < -1.737
Don't really understand the steps above. Is it possible for someone to explain it?

[Yertle the Turtle]

Thanks Bell9565 and Sine
Had another question
solve 2^x < 0.3
worked solutions from textbook
log₂(0.3) < x
-1.737 > x
x < -1.737
Don't really understand the steps above. Is it possible for someone to explain it?

You use your understanding that a^b=c is the same as b=loga(c) (or log of c to the base a).
Therefore 2^x<0.3 is the same as log2(0.3)<x
So you find log2(0.3) and substitute into the equation to find the inequality for x

[snowisawesome]

You use your understanding that a^b=c is the same as b=loga(c) (or log of c to the base a).
Therefore 2^x<0.3 is the same as log2(0.3)<x
So you find log2(0.3) and substitute into the equation to find the inequality for x
[/quote


Do you know why
log2(0.3) < x becomes -1.737 > x
Thanks

[lzxnl]

Thanks Bell9565 and Sine
Had another question
solve 2^x < 0.3
worked solutions from textbook
log₂(0.3) < x
-1.737 > x
x < -1.737
Don't really understand the steps above. Is it possible for someone to explain it?

You've written the signs incorrectly here.

[Shadowxo]

Do you know why
log2(0.3) < x becomes -1.737 > x
Thanks

Because log₂0.3 = -1.737 (can plug into calculator to get the value).

Also when solving inequalities, if you're not doing something standard (adding subtracting multiplying etc) then sometimes you have to flip the sign. A bit rusty on this though.
I'd solve as though they're equal and then see whether it needs to be less than or greater than by seeing what happens to the original inequality if x increases or decreases. If the original function (2^x in this case) increases as x increases, you keep the sign the same and the x on the same side as the function originally was in, if it's the opposite then you flip the sign.


Note: Someone can probably explain a lot better than me and don't worry if you don't understand it, just pointing out you have to worry about the direction of the sign

Edit: didn't see that Marvin mentioned the inequality, oops. You can also substitute a close value in to see which direction it goes

[Sine]

I'd solve it the exact same way as Shadowxo, the only difference would be that I'd graph it to determine the direction of the sign. 

Unless the inequality is linear it is much safer to not try to solve the inequality directly since it can get confusing on when to flip the sign or when you can leave it.

[lzxnl]

I'd solve it the exact same way as Shadowxo, the only difference would be that I'd graph it to determine the direction of the sign. 

Unless the inequality is linear it is much safer to not try to solve the inequality directly since it can get confusing on when to flip the sign or when you can leave it.

The general rule is, if you apply an increasing function to both sides, the inequality sign stays the same. For instance, x^3 < 1 -> x < 1 is valid because cube roots are monotonically increasing.

If you apply a decreasing function to both sides, the inequality sign flips. For instance, -x < 1 -> x > -1 because applying the function f(x) = -x (i.e. multiplication by -1) is a decreasing function.

[pha0015]

How would you use integration by recognition to solve anti differentiation for log?