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Suppose that a, b, and c are real numbers greater than 1. Then the value of $$\dfrac{1}{1+\log_{a^2b}\dfrac{c}{a}} + \dfrac{1}{1+\log _{b^2c}\dfrac{a}{b}} + \dfrac{1}{1+\log_{c^2a}\dfrac{b}{c}}$$ is .............
Correct Answer: 3
$$\dfrac{1}{\log_{a^2b}a^2b+\log_{a^2b}\dfrac{c}{a}} + \dfrac{1}{\log_{b^2c}b^2c+\log _{b^2c}\dfrac{a}{b}} + \dfrac{1}{\log_{c^2a}c^2a+\log_{c^2a}\dfrac{b}{c}}$$
$$\dfrac{1}{\log_{a^2b}a^2b\times\dfrac{c}{a}} + \dfrac{1}{\log _{b^2c}b^2c\times\dfrac{a}{b}} + \dfrac{1}{\log_{c^2a}c^2a\times\dfrac{b}{c}}$$
$$\dfrac{1}{\log_{a^2b}abc} + \dfrac{1}{\log _{b^2c}abc} + \dfrac{1}{\log_{c^2a}abc}$$
$$\log_{abc}a^2b+\log_{abc}b^2c+\log_{abc}c^2a$$
$$\log_{abc}a^3b^3c^3$$
$$\log_{abc}{abc}^3$$
$$=3$$
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