Power of variables!

Calculus Level 3

If 2 x = 3 y = 1 2 z 2^{x} = 3^{y} = 12^{z} then find the value of k k and q q in the equation below:

1 z = 1 k + q x \large \frac 1z = \frac 1k + \frac qx

y 2 , 1 y^{2} ,1 y , 1 y , 1 1 2 , 1 \dfrac{1}{2} , 1 y , 2 y , 2 2 , 1 2 , 1 x 2 , 2 x^{2} , 2 1 , 1 1 , 1 y z , 2 yz , 2

Arpan Ray by Arpan Ray , 17, India

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1 solution

Arpan Ray
May 31, 2017

2^ x x = 3^ y y = 12^ z z

Let it be equal to k k . Then; 2^ x x = 3^ y y = 12^ z z = k k

So, this means

2^ x x = k k 3^ y y = k k 12^ z z = k k
2 = k k ^ 1 x \frac{1}{x} 3 = k k ^ 1 y \frac{1}{y} 12 = k k ^ 1 z \frac{1}{z}

and we know that 12 = 2^2 X 3

So, 12 = k k ^ 1 z \frac{1}{z}

=> 2^2 X 3 = k k ^ 1 z \frac{1}{z}

[Instead of 2, we will put k k ^ 1 x \frac{1}{x} ]

[Instead of 3, we will put k k ^ 1 y \frac{1}{y} ]

( k k ^ 1 x \frac{1}{x} )^2 X k k ^ 1 y \frac{1}{y} = k k ^ 1 z \frac{1}{z}

k k ^ 2 x \frac{2}{x} X k k ^ 1 y \frac{1}{y} = k k ^ 1 z \frac{1}{z}

k k ^ 2 x \frac{2}{x} + 1 y \frac{1}{y} = k k ^ 1 z \frac{1}{z}

[Same base will be cut]

2 x \frac{2}{x} + 1 y \frac{1}{y} = 1 z \frac{1}{z}

And as we can see, it is same as the next equation q x \frac{q}{x} + 1 k \frac{1}{k} = 1 z \frac{1}{z}

With 2 in place of q q and y y in place of k k .

Hence k k = y y ; q q = 2

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