VCE Methods Question Thread!

[snowisawesome]

Yes - it's the same as (sin x)² = (sin(x))². The extra sets of brackets are there to make it less ambiguous.

Thanks
Am I right in thinking that this rule would apply to any situation?

[Bri MT]

Thanks
Am I right in thinking that this rule would apply to any situation?

Yes

[snowisawesome]

Yes

Thanks
had another question
If you had the equation x+2
and it was dilated by a factor of 1/2 from the y axis
then would the new equation become 2(x+2) = 2x+4
or would it just become 2x+2
Thanks
Also, for log₉(3xy) = 1.5
if we had to express y in terms of x
then is it correct to do
3xy=9^(1.5)
so y = ((9^(1.5))/3x
but the answer had a different working out and got the answer as y = 9/x
is there a way to work out 9^(1.5) without cas?
Thanks again

Express y in terms of x for the following equations
log₄y = -2+2log₄x
so I did log₄y = -2log₄4 + 2log₄x
So i got log₄y = -log₄(4^2) + log₄(x^2)
So then I did y = (4^2)(x^2) = 16*x^2 = 16x^2 due to adding the logs as shown in bold above
But the answer said log₄(x^2) - log₄(4^2) = log₄
((x^2))/(16)
So how do you determine weather it's -log₄(4^2) + log₄(x^2) or
log₄(x^2) - log₄(4^2)

Which order is correct and why?
Thanks

[lzxnl]

Thanks
had another question
If you had the equation x+2
and it was dilated by a factor of 1/2 from the y axis
then would the new equation become 2(x+2) = 2x+4
or would it just become 2x+2
Thanks
Also, for log₉(3xy) = 1.5
if we had to express y in terms of x
then is it correct to do
3xy=9^(1.5)
so y = ((9^(1.5))/3x
but the answer had a different working out and got the answer as y = 9/x
is there a way to work out 9^(1.5) without cas?
Thanks again

Express y in terms of x for the following equations
log₄y = -2+2log₄x
so I did log₄y = -2log₄4 + 2log₄x
So i got log₄y = -log₄(4^2) + log₄(x^2)
So then I did y = (4^2)(x^2) = 16*x^2 = 16x^2 due to adding the logs as shown in bold above
But the answer said log₄(x^2) - log₄(4^2) = log₄
((x^2))/(16)
So how do you determine weather it's -log₄(4^2) + log₄(x^2) or
log₄(x^2) - log₄(4^2)

Which order is correct and why?
Thanks

As for your second question,

always. Both orders you gave are correct because addition is commutative (can do in the opposite order too) by definition.

[snowisawesome]

As for your second question,

always. Both orders you gave are correct because addition is commutative (can do in the opposite order too) by definition.

But woudn't it make a difference it I did
((x^2)/(16)) vs doing ((x^2)(16))
Also, do you know how to do my first question highlighted in bold?
If you had the equation x+2
and it was dilated by a factor of 1/2 from the y axis
then would the new equation become 2(x+2) = 2x+4
or would it just become 2x+2
Thanks

[VanillaRice]

But woudn't it make a difference it I did
((x^2)/(16)) vs doing ((x^2)(16))

The commutative property of addition means that it can be done in any order.
The example you have provided suggests that multiplication and division are the same - they are not. (However, mutiplcation is also commutative).

Also, do you know how to do my first question highlighted in bold?
If you had the equation x+2
and it was dilated by a factor of 1/2 from the y axis
then would the new equation become 2(x+2) = 2x+4
or would it just become 2x+2
Thanks

Since dilations from the y-axis changes x-values, the resulting function would be y = 2x + 2.

Hope this helps

[snowisawesome]

The commutative property of addition means that it can be done in any order.
The example you have provided suggests that multiplication and division are the same - they are not. (However, mutiplcation is also commutative).
Since dilations from the y-axis changes x-values, the resulting function would be y = 2x + 2.

Hope this helps

What do you mean by the commutative property of addition?

[TheAspiringDoc]

What do you mean by the commutative property of addition?

a+b+c = c+a+b
E.g. 5 + (-3) = -3 + 5

[snowisawesome]

a+b+c = c+a+b
E.g. 5 + (-3) = -3 + 5

Thanks
Do you know how to make 64^-1 into a base of 2
like, 64^-1 = 2^x
Do you know how to find x?
Thanks again

[Yertle the Turtle]

Thanks
Do you know how to make 64^-1 into a base of 2
like, 64^-1 = 2^x
Do you know how to find x?
Thanks again

64=2^6
=>(2^6)^-1=2^x
Use your index laws to find that:
2^-6=2^x
=>x=-6

[Shadowxo]

Thanks
Do you know how to make 64^-1 into a base of 2
like, 64^-1 = 2^x
Do you know how to find x?
Thanks again

64 = 2⁶
64^-1 =(2⁶)^-1 = 2^-6
x=-6
Just beaten by Marvin

[snowisawesome]

Thanks
Shadowxo and Marvin

Had another question
Does it matter whether you say x = log_e(9)+2 or x = 2log_e(3)+2
Thanks

[VinnyD]

Thanks
Shadowxo and Marvin

Had another question
Does it matter whether you say x = log_e(9)+2 or x = 2log_e(3)+2
Thanks

I think it would depend on the question There are some questions that require you to write your answer in the form alog(b)+c and others that do not. I personally tend to put a coefficient out in front of the log when i can.
Hope this helps!

[snowisawesome]

I think it would depend on the question There are some questions that require you to write your answer in the form alog(b)+c and others that do not. I personally tend to put a coefficient out in front of the log when i can.
Hope this helps!

Thanks
I asked this question yesterday on this thread but i'm still confused with this question so would it be possible for you to explain it?

Express y in terms of x for the following equations
log4y = -2+2log4x
so I did log4y = -2log44 + 2log4x
So i got log4y = -log4(4^2) + log4(x^2)
So then I did y = (4^2)(x^2) = 16*x^2 = 16x^2 due to adding the logs as shown in bold above
But the answer said log4(x^2) - log4(4^2) = log4
((x^2))/(16)
So how do you determine weather it's -log4(4^2) + log4(x^2) or
log4(x^2) - log4(4^2)

Which order is correct and why?
Thanks
Thanks again

[VinnyD]

Thanks
I asked this question yesterday on this thread but i'm still confused with this question so would it be possible for you to explain it?

Express y in terms of x for the following equations
log4y = -2+2log4x
so I did log4y = -2log44 + 2log4x
So i got log4y = -log4(4^2) + log4(x^2)
So then I did y = (4^2)(x^2) = 16*x^2 = 16x^2 due to adding the logs as shown in bold above
But the answer said log4(x^2) - log4(4^2) = log4
((x^2))/(16)
So how do you determine weather it's -log4(4^2) + log4(x^2) or
log4(x^2) - log4(4^2)

Which order is correct and why?
Thanks
Thanks again

Well you can't use the addition law for logs because one log term is negative and one is positive. If you are going to add the logs it should just be 'loga(b) by itself'. In our case it is " -loga(b)". Because of this, we have to take the negative up as a power of b; only then can we use the addition rule. When you used the rule before to get 16x^2, you disregarded the negative, which you cannot do. It would make more sense to use the subtraction law because there is a negative. I guess the point is you cannot add logs when there is a negative sign in front of it. You have to first take the negative as a power and then you can add the two logs. A negative sign makes quite a bit of difference

Hope it makes sense!