VCE Methods Question Thread!

[alchemy]

Been doing fine with all other questions but how do i do this log question? Someone take me through the steps?
Express in simplest form: 
log₃(27)+1
log₄(16)+3

If you feel more comfortable working with pro numerals you can try this:
log₃(27)  = x-1
3^x-1=27
3^x-1=3³
x-1=3
x=4

After this, to get to the answer you mentioned, you can do log₃(y) = 4 and solve for y. Therefore, y=3⁴=81. Write it out completely and you get log₃ (81) = 4.
Tbh, there's no difference between this and the method suggested by psyxwar. Just preference I guess, and *possibly* a better way of understanding the situation when working with very large numbers.

[~V]

The first one is really clear now but the second one, where did 1024 come from..?

[Blondie21]

log₄(16)+3

log₄(16)+3
log₄(16)+3log₄(4) (because 1 = log₄(4) as mentioned before)
log₄(16)+log₄(4)³
log₄(16)+log₄(64)
log₄(16*64)
log₄(1024)
Then to find what this is equal to...
log₄(4)⁵
5log₄(4)
=5
.:. log₄(1024)=5

[ETTH96]

Hey guys, I need help with this question:

for what values of k and n does the following set of simultaneous equations have an infinite number of solutions?

kx+4y=2n
2x+(k+2)y=-1

Do i let them equal each other or... what does the 'infinite number of solutions' tell me?

Thanks

[b^3]

If you have infinite number of solutions then both lines are basically the same line. So that is their gradients are the same and their intercept is the same. You can do this one of two ways, using matrix methods or by rearranging the equation and then comparing the gradient and intercept.

So using the second method we have:


Now we want the gradient and intercept to be the same, so we can equate them forming two simultaneous equations that we can solve for and .

Spoiler

[ETTH96]

If you have infinite number of solutions then both lines are basically the same line. So that is their gradients are the same and their intercept is the same. You can do this one of two ways, using matrix methods or by rearranging the equation and then comparing the gradient and intercept.

So using the second method we have:


Now we want the gradient and intercept to be the same, so we can equate them forming two simultaneous equations that we can solve for and .

Spoiler

Thanks so much, I understand that now!

What if it says "unique solution" or "no solution"? What do I do then?

The question is:

Consider the linear simultaneous equations:

kx-4y=2
3x-(k+4)y=k+1

where k is a real constant

i) Find the value(s) of k, for which there is a unique solution
ii) Find the value(s) of k, for which there is no solution

How do I go about doing this?? Thanks

[Phy124]

Thanks so much, I understand that now!

What if it says "unique solution" or "no solution"? What do I do then?

The question is:

Consider the linear simultaneous equations:

kx-4y=2
3x-(k+4)y=k+1

where k is a real constant

i) Find the value(s) of k, for which there is a unique solution
ii) Find the value(s) of k, for which there is no solution

How do I go about doing this?? Thanks

Chucking this in a spoiler tag because it's quite large

A unique solution is when the two lines intersect on one occasion, this occurs when the determinant is not equal to zero.

So:









Hence represent the solutions for when there are unique solutions.



There is no solution when the two lines are parallel. This occurs when the determinant is zero.

So:

Same business as before:











If :









These two lines are parallel to each other and hence represents the solution for no solutions.



If :









These two lines are the same and hence represents the solution for infinitely many solutions.

[b^3]

Thanks so much, I understand that now!

What if it says "unique solution" or "no solution"? What do I do then?

The question is:

Consider the linear simultaneous equations:

kx-4y=2
3x-(k+4)y=k+1

where k is a real constant

i) Find the value(s) of k, for which there is a unique solution
ii) Find the value(s) of k, for which there is no solution

How do I go about doing this?? Thanks

If the gradients of two lines are the same then we can never have a unique solution (as if the gradients are the same the two lines are parallel and we either have the same line twice or two separate parallel lines that don't intersect).

So for unique solutions we need: - the gradients of the two lines to be different, . So rearrange the two equations and 'equate' the gradients, but use a (not equal to) sign instead.

The no solution case is where we have two parallel lines that are not the same line (as if they were we'd get infinite solutions rather than no solutions). So what we need to do is find when the two lines have the same gradient but have a different intercept (as if the intercept is the same then we have the same line, which we don't want).
So:
 - Rearrange the two lines into form.
 - Equate the two gradients,
 - The two intercepts are not equal,

EDIT: Beaten, although note Phy used the Matrix method (which is shorter most of the time, I always seemed to prefer the rearrange and compare method though, just depends on which works better for you).

[Nato]

Hey guys, i was wondering why do we let the determinant=0 when trying to find unique/no solutions and not equal to zero when trying to find unique solutions? How can the determinant of a matrix, enable us to figure this out?

[b^3]

If you have two linear equations, and then we can represent the system of linear equations with a matrix equation.

Now if we were to try and solve this matrix equation for (which represents ) then we would try to first take the inverse of and multiply both sides by it, since we are trying to get on it's own (we need to get rid of the on the LHS side, and we can do that by 'hitting' it with it's inverse from the left to give us the identity matrix , which effectively leaves us with whatever matrix we multiply that by).

Now this will only work if we have a unique solution, that is we can only find an inverse of if the system has a unique solution. For an inverse of to exist, the determinate of , or must be non zero.
So this means for:
 - a unique solution,
 - no/infinite solutions,
For no and infinite solutions case you will then need to be able to pick between the two by substituting your values found back into the equations and seeing if the come out as the same equation or not.

Hope that helps

EDIT: tl;dr, it comes down to the fact that you need the inverse of to exist or not exist, which is then linked to the determinate of the matrix being zero or non-zero.

[Bluegirl]

Could someone help me please?
I'm stuck.
If log a(x) =0.3 and log a(y) =0.2 find log a(xy)

[noah the lettuce]

Hey guys, I have a few questions...
Find exact solution(s) to each of these
1.    16-(x+3/x)^2=0
2.    x^4-x^3+x-1=0
3.    1+√(x+3)=-x
4.    √(5-x)-√x=√(2x-1)

[Nato]

Could someone help me please?
I'm stuck.
If log a(x) =0.3 and log a(y) =0.2 find log a(xy)

Hey bluegirl,

try and think about what equals, if you broke it up. It corresponds with one of the log laws.

[Nato]

Hey guys, I have a few questions...
Find exact solution(s) to each of these
1.    16-(x+3/x)^2=0
2.    x^4-x^3+x-1=0
3.    1+√(x+3)=-x
4.    √(5-x)-√x=√(2x-1)

here are a few suggestions
for 1. start by letting then solve for a. After you receive the value(s) for , make it equal to , to solve for

for 2. try and use the factor theorem, so subbing in values that make equal to . This will help you in finding a factor. Which you can then use long division to find other factors, and hence the solution(s)

for 3. rearrange the equation to look like: . Now you can square both sides of the equation to 'eliminate' the square root sign. After a bit of expanding and rearranging you will be left with a 'normal' quadratic, which you can then solve for

for 4. kinda similar to 3. you could start off by squaring both sides:
so on the right hand side the square root will be eliminated, and on the left hand side, i would advise to expand it (using difference of perfect squares). this expansion will resulting in a square root term, but after a bit of rearranging and squaring both sides, you can find the value(s) for x.

Let me know, if that didn't make sense haha cheers!

[b^3]

Hey guys, I have a few questions...
Find exact solution(s) to each of these
1.    16-(x+3/x)^2=0
2.    x^4-x^3+x-1=0
3.    1+√(x+3)=-x
4.    √(5-x)-√x=√(2x-1)

We'll give you hints rather than doing all the working for you.
1. Multiply both sides by to get rid of the you have in one of the denominators.
2. Group the terms and , take a factor out of the first which should leave you with both groupings that have another factor in common, then factorise this factor out to give you something that you can work with via the null factor law.
3. You need to first get rid of the square root, so to do this we normally square both sides, but as we have , if you square the left side you'll still get a square root when you expand it. So we first need to get the square root on it's own by moving the across, then square both sides and expand out to give you a quadratic. ALso take note of any domain restrictions that arise from having the square root in there (remember anything 'underneath' the square root has to be equal to or greater than zero, which sometimes restricts your solutions).
4. You'll need to do the rearranging and squaring both sides twice for this one, as the first time you'll still end up with a square root left over. Again be careful with domain restrictions from the square root.

Spoiler

4.

Check domain restrictions,

Which means that is not a valid solution.

EDIT: Beaten.