Question

If 5x

= 30-y

= 6z

, then what is the value of (xy + yz + zx)/xyz ?​

Answer 1

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Answer 2

$\frac{xy+yz+zx}{xyz} = 0$

Solution:

Given that,

$5^x = 30^{-y} = 6^z$

From given,

$5^x = 30^{-y}\\\\5^x = (5 \times 6)^{-y}\\\\5^x = 5^{-y} \times 6^{-y}\\\\On\ rearranging\ we\ get\\\\\frac{5^x}{5^{-y}} = 6^{-y}\\\\Therefore\\\\5^{x +y} = 6^{-y}$

Take power -1/y on both sides

$5^{\frac{x+y}{-y}} = 6^{\frac{-y}{-y}}\\\\5^{\frac{x+y}{-y}} = 6 --------- eqn\ 1$

Similarly,

$30^{-y} = 6^z\\\\(5 \times 6)^{-y} = 6^z\\\\5^{\frac{-y}{y+z}} = 6 ------ eqn 2$

From eqn 1 and eqn 2

$5^{\frac{x+y}{-y}}= 5^{\frac{-y}{y+z}}\\\\We\ know\ that\\\\a^m = a^n\ then\ m = n\\\\\frac{x+y}{-y} = \frac{-y}{y+z}\\\\On\ cross\ multiplying\ we\ get\\\\y^2 = (x+y)(y+z)\\\\y^2 = xy +xz +y^2 + yz\\\\xy + xz + yz = 0\\\\Therefore\\\\\frac{xy + xz+yz}{xyz} = 0$

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