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Preliminary mathematics marathon (2 Viewers)

Gussy Booo

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Hi, I was just wondering if this solution is entirely correct.

Doesn't this imply that for any value for 'a' that the answer is '3a+2'? However doesn't this violate the first condition as a^2=a+2 only has two solutions?
Any value of "a", the answer is 3a+2 ......??
a^3 = 3a+2 my friend, not a.

I got this question from The Australian Maths Competition :D, hehe :) .
 

Gussy Booo

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Wow! Very nice Drongoski. Very creative how you used logs.

But here is an alternative solution:

Let 2^x = 5^y = 10^z = A

Then: 2 = A^(1/x) , 5 = A^(1/y) , 10 = A(1/z)

Now, 10 = 2 x 5.

.'. A(1/z) = A^(1/x) x A^(1/y) = A^(1/x + 1/y)

Now, all the "front values" are the same.

So: 1/z = 1/x + 1/y = (x+y)/xy
Therefore: z = xy/(x+y) #
 

Drongoski

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Very nice.
I like the change of base part.

But he said you didn't need to use log, didn't he?
I didn't see that; but logs & indices(exponents) are closely linked. What you can do via logs, you should be able to do via corresponding exponential equations. I have not tried that. Doing via logs was already quite a challenge, for me, anyway.
 

Gussy Booo

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Very nice.
I like the change of base part.

But he said you didn't need to use log, didn't he?
Well, that's the beauty of maths :) . There are many ways of retrieving an answer.
Look above for an alternative solution.
 

hscishard

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Wow! Very nice Drongoski. Very creative how you used logs.

But here is an alternative solution:

Let 2^x = 5^y = 10^z = A

Then: 2 = A^(1/x) , 5 = A^(1/y) , 10 = A(1/z)

Now, 10 = 2 x 5.

.'. A(1/z) = A^(1/x) x A^(1/y) = A^(1/x + 1/y)

Now, all the "front values" are the same.

So: 1/z = 1/x + 1/y = (x+y)/xy
Therefore: z = xy/(x+y) #
Makes sense. Will never be able to think like that in an exam.
 

Gussy Booo

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Makes sense. Will never be able to think like that in an exam.
That's why past papers are the key to success :D.
You've seen the question now. If you see a similar question in an exam, you've increased your chances to solve it by at least 60%.
 

Gussy Booo

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More Questions :D !!!

1) If 2^(2p).3^(p) = 12^(x), find p in terms of x and hence find p such that 2^(2p).3^(p)=144

2) If sinA = 2sinB and tanA = 3tanB, find A and B to the nearest degree, given that A and B are acute angles

Have fun!
 

Drongoski

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Wow! Very nice Drongoski. Very creative how you used logs.

But here is an alternative solution:

Let 2^x = 5^y = 10^z = A

Then: 2 = A^(1/x) , 5 = A^(1/y) , 10 = A(1/z)

Now, 10 = 2 x 5.

.'. A(1/z) = A^(1/x) x A^(1/y) = A^(1/x + 1/y)

Now, all the "front values" are the same.

So: 1/z = 1/x + 1/y = (x+y)/xy
Therefore: z = xy/(x+y) #
Wow! Although quite good at indices, I'd not have thought this one out myself. It's so elegant.
 

AAEldar

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More Questions :D !!!

1) If 2^(2p).3^(p) = 12^(x), find p in terms of x and hence find p such that 2^(2p).3^(p)=144

2) If sinA = 2sinB and tanA = 3tanB, find A and B to the nearest degree, given that A and B are acute angles

Have fun!
For 1:



Hmmm...On looking at this I don't think it makes sense :S
 
Last edited:

hscishard

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More Questions :D !!!

1) If 2^(2p).3^(p) = 12^(x), find p in terms of x and hence find p such that 2^(2p).3^(p)=144

2) If sinA = 2sinB and tanA = 3tanB, find A and B to the nearest degree, given that A and B are acute angles

Have fun!
For Q2.
It's funny.
For the first part A and B could be 0
But for the Tan part, A can be 60 and B can be 30. Or 0 :D
 

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