Question 32

If $$2^{x} = 4^{y} = 8^{z}$$ and $$(\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z}) = \frac{24}{7}$$ value of z is

Solution

Since x , y and z are expressed as powers of 2, 4 and 8, they have a relationship of x= 2y and x = 3z

So, if x = 6k, y = 3k and z= 2k

Substituting in $$(\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z}) = \frac{24}{7}$$ gives

1/4k = 24/7

So, k = 7/96

z = 2k = 7/48

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