TheAspiringDoc's Math Thread

[cosine]

That's my point - what you've written is only true for a=0,2. For you to state that (which you did), then it HAS to be true for all a. It's not. What you've written is wrong.

The book stated that too, and aspiringdoc wanted to see how to get it into that form. My original answer is at it is, but because I was being generous and decided to enlighten aspirigdoc in how to get it into the book's form, I decided too. Not sure where you are trying to get at man?

[IntelxD]

The book stated that too, and aspiringdoc wanted to see how to get it into that form. My original answer is at it is, but because I was being generous and decided to enlighten aspirigdoc in how to get it into the book's form, I decided too. Not sure where you are trying to get at man?

Can I please just interject so we can put an end to this nonsense? Eulerfan is trying to point out that   isn't equal to  . Saying they are equal is akin to saying 1/(3*6) is equal to (1/3)*6. However, as even a broken clock is right twice a day, there are certain values of a which hold true for the proposed equality. Therefore, if you chose to simplify in this manner, you must restrict the values of a so that your simplification is valid.

[cosine]

Can I please just interject so we can put an end to this nonsense? Eulerfan is trying to point out that   isn't equal to  . Saying they are equal is akin to saying 1/(3*6) is equal to (1/3)*6. However, as even a broken clock is right twice a day, there are certain values of a which hold true for the proposed equality. Therefore, if you chose simplify in this manner, you must restrict the values of a so that your simplification is valid.

I would usually ignore banter like this, but I am not seeing the point of your unnecessary needs to invade..

Simplify: log(x^2)
= 2log(x)

Pretty sure you would not include the x cannot equal 0 after it, as we are not worried about the values that can go into it in this certain context, but rather more worried about how we can simplify it. So, are you saying that the above expression is wrong?

[cosine]

Simplify: log(x^2) does not equal
 2log(x)

 log(x)^2
= 2log(x)

Incorrect.

(log(x))^2 = log(x) * log(x)

[keltingmeith]

I would usually ignore banter like this, but I am not seeing the point of your unnecessary needs to invade..

Simplify: log(x^2)
= 2log(x)

Pretty sure you would not include the x cannot equal 0 after it, as we are not worried about the values that can go into it in this certain context, but rather more worried about how we can simplify it. So, are you saying that the above expression is wrong?

Yes, actually. That simplification only makes sense if x is positive, and you have to note that, or you will run into problems later. Try solving log_10(x^2)=20 using that method and see what happens, then compare your answer to a graph.

[TheAspiringDoc]

Solve the following equation, expressing the value of x with a rational denominator

 Thanks

[Callum@1373]

Solve the following equation, expressing the value of x with a rational denominator

 Thanks

Then just multiply by to get a rational denominator

It's just using the basic algebra rules 

[TheAspiringDoc]

Then just multiply by to get a rational denominator

It's just using the basic algebra rules 

but then we end up with rather than ?

[StupidProdigy]

but then we end up with rather than ?

multiplying by root 3 is just rationalising, which doesn't actually change the expression value, meaning you're basically just writing the answer in a different way even though it's the same value. Have a look into surd rationalisation a bit more is my suggestion

[cosine]

Solve the following equation, expressing the value of x with a rational denominator

 Thanks

AspiringDoc, between the last three lines, when I multiplied by sqrt3, it was because I was rationalising the denominator, i.e. expressing it in a more appropriate way, it is basically technically the same answer.

Edit: Beaten, but ill leave the working out for ya.

[TheAspiringDoc]

ahh.. so we're multiplying both the numeraator and the denominator of the fraction by  , just not the x. That makes a lot more sense - thanks guys

[TheAspiringDoc]

make 'b' the subject if
my book says 2A-ah/h this is wrong, right?
Cheers

[LoadedWithPotatoes]

The book's answer is correct. They expanded the bracket, eg

->
2A = (a+b)h
2A= ah+bh
(2A-ah)/h = b

If you did 2A/h - a = b instead that would be the same thing as the book's answer

[TheAspiringDoc]

Hi guys!
So I thought since this is my math thread, I can also add worked solutions to problems I thought were cool so other people can learn from them and if I got it wrong then someone might pick me up on it (if anyone still ever reads this thread!) 

So here's the first one:

Three students attempt to solve a mathematical puzzle independently of each other. The first student has a chance of 0.5 of solving the puzzle, the second student has a chance of 0.6 of solving the puzzle, while the third student has a chance of 0.7 of solving the puzzle.   

(a) What is the probability that all three solve the puzzle?
Pr(all solve puzzle) = 0.5x0.6x0.7
= 21/100 or 0.21 or 21%
 
(b) What is the probability none of the students solve the puzzle?
Pr(none solve puzzle) = (1-prob. studentA solves puzzle) x (1- prob student B solves puzzle) x (1-prob. student C solves puzzle)
=(1-0.5)x(1-0.6)x(1-0.7)
=0.5x0.4x0.3
=3/50 or 0.06 or 6%

Thanks!!

[TheAspiringDoc]

Howdy!!

Another worked solution (I hope!!)

Find the units digit of

Okay, so whilst this may initially look like it needs a supercomputer and simply isn't solvable with a handheld calculator, the fact is that it actually is.

So lets establish one thing:
we only care about the units digits ofwhen adding them together to find their units digits.

So, to find the units digit for is surprisingly easy. What we notice when raising 6^x   (when x isn't 0, that is) is that the units digit is always 6. For example:

6^1 = 6
6^2 = 36
6^3 = 216
6^4 = 1296
6^5 = 7776

etc.

So we now have:
find the units digit of

For the other two, we are going to have to use a somewhat trickier method, as I will now demonstrate on
When we raise 7 to any power, we see a pattern, that is:
7^0 = 1
7^1 = 7
7^2 =49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801
etc.

the pattern becomes clear when we look at the units digit of each one:
7^0 = 1
7^1 = 7
7^2 =9
7^3 = 3
7^4 = 1
7^5 = 7
7^6 = 9
7^7 = 3
7^8 = 1

etc.
So they are cycling through the numbers 1,7,9 and 3.
and there are four numbers in this cycle, and the 52 in 7^52 is a multiple of four, so it will have the first number of the cycle: 1.

So we now have:
find the units digit of

Doing the same for , we notice that it cycles as well
Units digits:

8^0 = 1
8^1 = 8
8^2 = 4
8^3 = 2
8^4 = 6
8^5 = 8
8^6 = 4

etc.

So our cycle is:
8,4,2,6.
Which also happens to have four components.
if the 22 in was a multiple of four (which it isn't) the units digit would be a 6. however it is two short of being a multiple of four (e.g. 24) so whatt we do is put our cycle in there, giving us:
(units digit of)
so now we just look at our cycle list (8,4,2,6) and look two places before 6, which is 4.
therefore (units digit of)

So we now have:
find the units digit of
which is 11, the units digit of which is 1.

Thus, the units digit of = 1! (the exclamation mark is there not as a factorial, but rather as an emoticon, by the way )

Hope you guys enjoyed B)

GAHHHH I HATE LATEX!!!!! (they're factorials btw)