If x, y, z are in G.P., $a^{x} = b^{y} = c^{z}$ , then-

l o g_{c} b = l o g_{a} c

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l o g_{1} c = l o g_{b} c

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l o g_{a} b = l o g_{c} b

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l o g_{b} a = l o g_{c} b

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Solution

The correct option is A $l o g_{c} b = l o g_{a} c$
$x l o g a = y l o g b = z l o g c = k (s a y)$
Also $y^{2} = x z$
$\frac{k^{2}}{(l o g b)^{2}} = \frac{k^{2}}{l o g a . l o g c}$
or $\frac{l o g a}{l o g b} = \frac{l o g b}{l o g c}$ or $l o g_{b} a = l o g_{c} b$

Suggest Corrections

Similar questions

l o g_{c} b + l o g_{b} c

y = l o g_{a} c + l o g_{c} a,

z = l o g_{b} a + l o g_{a} b

x y z = x^{2} + y^{2} + z^{2} - 4.

x = l o g_{b} a, y = l o g_{c} b, z = l o g_{a} c

\frac{1}{1 + {log}_{b} a + {log}_{b} c} + \frac{1}{1 + {log}_{c} a + {log}_{c} b} + \frac{1}{1 + {log}_{a} b + {log}_{a} c}

{log}_{b} a + {log}_{c} a + {log}_{a} b + {log}_{a} c + {log}_{b} c + {log}_{c} b = 3

a, b, c

\neq 1

a b c

\frac{1}{1 + {log}_{b} a + {log}_{b} c} + \frac{1}{1 + {log}_{c} a + {log}_{c} b} + \frac{1}{1 + {log}_{a} b + {log}_{a} c}

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