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Logarithms

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Question : 33 of 85

Marks: +1, -0

If

\; \frac{\log x}{a^{2}+ab+b^{2}} = \; \frac{\log y}{b^{2}+bc+c^{2}} = \; \frac{\log z}{c^{2}+ca+a^{2}}

x^{a-b} \cdot y^{b-c} \cdot z^{c-a} =

1
$-1$
0
2.

Solution:

Concept:

Use the common ratio

k

to express each logarithm and then compute the exponent sum.

Explanation:

Let the common value be

k

. Then:

\log x = k (a^2 + ab + b^2)

,

\log y = k (b^2 + bc + c^2)

,

\log z = k (c^2 + ca + a^2)

.

Consider

\log(x^{a-b} y^{b-c} z^{c-a}) = (a-b)\log x + (b-c)\log y + (c-a)\log z

.

Substitute:

k[(a-b)(a^2+ab+b^2) + (b-c)(b^2+bc+c^2) + (c-a)(c^2+ca+a^2)]

.

Each product is a difference of cubes:

(a-b)(a^2+ab+b^2) = a^3 - b^3

,

(b-c)(b^2+bc+c^2) = b^3 - c^3

,

(c-a)(c^2+ca+a^2) = c^3 - a^3

.

Sum:

(a^3 - b^3) + (b^3 - c^3) + (c^3 - a^3) = 0

.

Thus

\log(x^{a-b} y^{b-c} z^{c-a}) = 0

, so

x^{a-b} y^{b-c} z^{c-a} = 10^0 = 1

.

Answer:

1

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