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Question

Find the value of $$\dfrac{1}{1+\log_b a+\log_bc}+\dfrac{1}{1+\log_ca+\log_cb}+\dfrac{1}{1+\log_ab+\log_ac}$$.


A
2
loader
B
1
loader
C
0
loader
D
none of these
loader

Solution

The correct option is A $$1$$
Now,
$$\dfrac{1}{1+\log_b a+\log_bc}+\dfrac{1}{1+\log_ca+\log_cb}+\dfrac{1}{1+\log_ab+\log_ac}$$
$$=\dfrac{1}{\log_bb+\log_b a+\log_bc}+\dfrac{1}{\log_cc+\log_ca+\log_cb}+\dfrac{1}{\log_a a+\log_ab+\log_ac}$$
$$=\dfrac{1}{\log_b abc}+\dfrac{1}{\log_c abc}+\dfrac{1}{\log_aabc}$$ [ Using product rule of logarithtm]
$$=\dfrac{\log b}{\log abc}+\dfrac{\log c}{\log abc}+\dfrac{\log a}{\log abc}$$
$$=\dfrac{\log abc}{\log abc}$$
$$=1$$.

Mathematics

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