Logarithm relation

Algebra Level 4

{ x = log ⁡ 2 k ( k ) y = log ⁡ 3 k ( 2 k ) z = log ⁡ 4 k ( 3 k ) \large{\begin{cases} x = \log_{2k}(k) \\ y = \log_{3k}(2k) \\ z = \log_{4k}(3k) \end{cases}}

Establish a relation between x , y , z x , y , z , such that the above system of equations satisfies

x y + y z + z x = x y z xy + yz + zx = xyz None of these choices x y z + 1 = 2 y z xyz + 1 = 2yz x y + y z + z x = x y z + 1 xy + yz + zx = xyz + 1 x y + y z + z x = 2 x y z xy + yz + zx = 2xyz

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Akhil Bansal by Akhil Bansal , 22, India

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1 solution

By the change of base rule, we have that

x y z = log ⁡ ( k ) log ⁡ ( 2 k ) ∗ log ⁡ ( 2 k ) log ⁡ ( 3 k ) ∗ log ⁡ ( 3 k ) log ⁡ ( 4 k ) = log ⁡ ( k ) log ⁡ ( 4 k ) = log ⁡ 4 k ( k ) . xyz = \dfrac{\log(k)}{\log(2k)}*\dfrac{\log(2k)}{\log(3k)}*\dfrac{\log(3k)}{\log(4k)} = \dfrac{\log(k)}{\log(4k)} = \log_{4k}(k).

Next, note that

2 y z = 2 ∗ log ⁡ ( 2 k ) log ⁡ ( 3 k ) ∗ log ⁡ ( 3 k ) log ⁡ ( 4 k ) = 2yz = 2*\dfrac{\log(2k)}{\log(3k)}*\dfrac{\log(3k)}{\log(4k)} =

2 ∗ log ⁡ ( 2 k ) log ⁡ ( 4 k ) = 2 log ⁡ 4 k ( 2 k ) = log ⁡ 4 k ( 4 k 2 ) = log ⁡ 4 k ( 4 k ) + log ⁡ 4 k ( k ) = 1 + x y z . 2*\dfrac{\log(2k)}{\log(4k)} = 2\log_{4k}(2k) = \log_{4k}(4k^{2}) = \log_{4k}(4k) + \log_{4k}(k) = 1 + xyz.

Thus the correct option is x y z + 1 = 2 y z . \boxed{xyz + 1 = 2yz}.

(The other equation options can be quickly ruled out by plugging in k = 1 k = 1 .)

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Nice and fast solution!!!:-)

José Bezerra Carvalho Júnior - 5 years, 8 months ago

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