If $a^{x} = b^{y} = c^{z}$ & $y^{2} = x z$ , then the value of ${log}_{b} a$ will be

{log}_{c} b

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{log}_{b} c

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a {log}_{c} b

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None of these

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Solution

The correct option is A ${log}_{c} b$
$L e t a^{x} = b^{y} = c^{z} = P {log}_{a} P = x {log}_{b} P = y {log}_{c} P = z y^{2} = x z S o {log}_{b} P \cdot {log}_{b} P = {log}_{a} P {log}_{c} P Assuming base P, we get \frac{log P}{log b} \times \frac{log P}{log b} = \frac{log P}{log a} \frac{log P}{log c} So we can write \frac{1}{log b} \frac{1}{log b} = \frac{1}{log a} \frac{1}{log c} \frac{log a}{log b} = \frac{log b}{log c}$

${log}_{b} a = {log}_{c} b$

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Similar questions

y^{2} = x z

a^{x} = b^{y} = c^{z}

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

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.

\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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