VCE Methods Question Thread!

[SE_JM]

THANK YOU!!!!!!!
You have no idea how hard I tried to solve that...

I have one more question, if you would be so kind as to reply again

it is another logarithm/exponential question.

solve for x
2^(1/2)*(x+2) = 3^(x-1)

Again, thank you

[keltingmeith]

THANK YOU!!!!!!!
You have no idea how hard I tried to solve that...

I have one more question, if you would be so kind as to reply again

it is another logarithm/exponential question.

solve for x
2^(1/2)*(x+2) = 3^(x-1)

Again, thank you

Same process as before, but this time a little less messy. First, get rid of those annoying exponents:



Now, clean, factor, simplify:

[cosine]

Just a quick question, where and how do you get In(a) from and what does it mean? Why do you apply it to everything in the equation?

[keltingmeith]

Just a quick question, where and how do you get In(a) from and what does it mean? Why do you apply it to everything in the equation?

It's just shorthand because we use the logarithm to the base e so often. In truth, you can use whatever base you want - most mathematicians (myself included) will use base e by default whenever it doesn't matter what base they use. This is probably because base e works very nice in calculus (as you will find out this year), but in truth is just something I picked up because they all did it.

[AirLandBus]

Can someone explain the square root and circular functions graphing shiz. Our 1/2 (Hineee) only covers the square root function and the circular function.

Thus,

What is the difference between:

f(x)=-(squareroot) of (4-x^2)

How is this only a half a circle? is it because it is a square root.

and

f(x)= squareroot of (x^2-9)

Whats the difference?
We never got taught it.

[lzxnl]

Can someone explain the square root and circular functions graphing shiz. Our 1/2 (Hineee) only covers the square root function and the circular function.

Thus,

What is the difference between:

f(x)=-(squareroot) of (4-x^2)

How is this only a half a circle? is it because it is a square root.

and

f(x)= squareroot of (x^2-9)

Whats the difference?
We never got taught it.

I'm going to show you the difference between y = sqrt(x^2 - 1) and y = sqrt(1-x^2) for simplicity
For the first one, the domain is |x|>=1 and the range is all non-negative reals. For the second one, the domain is |x|<=1 and the range is 0<=y<=1. That's one difference already.

Then, for the first equation, if we square both sides, we get, for y>=0, y^2 = x^2 - 1
x^2 - y^2 = 1 which is a hyperbola (covered in spesh)

For the second equation, if we square both sides, we get, for y>=0, y^2 = 1 - x^2 or x^2 + y^2 = 1
Which is the part of the unit circle with y>=0. AKA half a circle.
Why is it half a circle? In the original function, y is the square root of something. Square roots are never negative. Hence you only have the part of the circle that is above the x axis, which is half the circle.

[AirLandBus]

Ok, so if its just a square root of say (x^2+9), it will spring outwards from -3 and 3.
If its reflected in both the x and y directions, it will be a semi-circle ( U shaped),
If its reflected in the y axis it will be a upside down "u" semi circle?
When its the square root of a x to the power of 1, it just springs out one side.
Whats the go with a square root of (4+x^2) graph, it seems its "stretched out" quadratic.


Correct?
Anything to add?

[lzxnl]

Ok, so if its just a square root of say (x^2+9), it will spring outwards from -3 and 3.
If its reflected in both the x and y directions, it will be a semi-circle ( U shaped),
If its reflected in the y axis it will be a upside down "u" semi circle?
When its the square root of a x to the power of 1, it just springs out one side.
Whats the go with a square root of (4+x^2) graph, it seems its "stretched out" quadratic.


Correct?
Anything to add?

It's not a 'stretched out quadratic'. y = sqrt(x^2 + 4) is a hyperbola as it's the same as y^2 = x^2 + 4 for y>=0

[AirLandBus]

If its reflected in the y-axis, it will be 'U semi circle'
If its reflected in the x-axis, the 'semi circle' will be upside down (flipped in the x-axis).

For , there are no possible x-values but there exists a y-value.



=

Now, for the x-value: let y=0

(get rid of the sq root by squaring the other side)





You cant take the square root of a negative number,  there are no possible x-values.

Now graph what you have, a y-value of (0,2) and thats it!

Sorry, i was referring to something else.
Say we have y=the squareroot of (x^2 +9) it will be the two "flares" from each value (-3 and 3).

Say we have y=the squareroot of (-x^2 +9) it will be the upside down U (reflections on the x axis).

Say we have y=the (-) squareroot of (-x^2 +9) it will be the U (reflections on the y and x axis).

And then the weird "quadratic" looking thingy is when its the square root of say (X^2 + a value). In which itself can be reflected on the x axis but if reflected on the y will turn into the upside dwon U shape.

So you cant have a square root function thats x^2 that will have the two kicks and be reflected on the x axis?

Think ive got it now. Thanks guys.

[AirLandBus]

I think ive got the jist of it now, just got a bit confused with previous stuff. Pidy we never got taught this.

[keltingmeith]

So everything in a logarithm equation, even as you did above: , you can convert to In(a) to make it easier? I mean how is 3 going to become in the form of In(a)? Could you give me an example I can try out, because my text book has nothing as hard as the above examples and it never mentioned anything about what you said.

Well, quick little things:

for all a>0

BUT, I didn't convert the equation - I just did an operation to both sides. Consider x+2=5.

If I want to solve for x, what I do is I need to get x by itself - so, I take away 2 from the LHS:

x+2-2 = 5

Except, what I wrote above is WRONG, so I compensate by taking a 2 from the RHS as well:

x+2-2=5-2
.: x=3

And so, I've solved the equation.

What I did was similar. Consider . The first thing I want to do is get x by itself, so I take a logarithm of the LHS to get rid of the power:



But, once again, this statement is wrong - so I need to counter this by doing it to the RHS as well:



And this is what I did earlier - I didn't /convert/ anything into log form. I just took the log of both sides.

[SE_JM]

Same process as before, but this time a little less messy. First, get rid of those annoying exponents:



Now, clean, factor, simplify:

So, you didn't expand the brackets initially...
What I did, was:

I first expanded the 2^(1/2)*(x+2) bit  so that would become 2^(x/2+1)
and then I simplified and yada yada, and then I got the wrong answer.

thank you so much for your help!

Could you explain why I couldn't expand this bracket first?

[lzxnl]

THANK YOU!!!!!!!
You have no idea how hard I tried to solve that...

I have one more question, if you would be so kind as to reply again

it is another logarithm/exponential question.

solve for x
2^(1/2)*(x+2) = 3^(x-1)

Again, thank you

So, you didn't expand the brackets initially...
What I did, was:

I first expanded the 2^(1/2)*(x+2) bit  so that would become 2^(x/2+1)
and then I simplified and yada yada, and then I got the wrong answer.

thank you so much for your help!

Could you explain why I couldn't expand this bracket first?

You can do that
2^(x/2 + 1) = 3^(x-1)
(x/2+1)*ln(2) = (x-1)*ln 3
x(1/2 ln 2 - ln 3) = -ln 3 - ln 2
x(ln 3 - 1/2 ln 2) = ln3 + ln 2
x = (ln 3 + ln2)/(ln 3 - 1/2 ln 2)

[SE_JM]

You can do that
2^(x/2 + 1) = 3^(x-1)
(x/2+1)*ln(2) = (x-1)*ln 3
x(1/2 ln 2 - ln 3) = -ln 3 - ln 2
x(ln 3 - 1/2 ln 2) = ln3 + ln 2
x = (ln 3 + ln2)/(ln 3 - 1/2 ln 2)

Ahh.. I think I may know what went wrong

Initially, I did get the right answer: x = log_e6/(log_e3-(1/2)*log_e2)  this was the correct answer I got at first
but then,
I simplified this

so it becomes this:

(log_e6)/(log_e2^(1/2))

and then I further simplified it to this:

(log_e6)/(log_e(3/(2^1/2))

When I put it in the calculator like this, I got a wrong answer, which was 8.84

But why was I wrong, if I only simplified it?

I must have done something really weird while "simplifying "this...

Can you spot where I made a mistake? I can't really tell

[Stevensmay]

Your simplification was correct, maybe a mistake entering into the calculator?
In this case simplification is not needed anyway.