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EXERCISE 6.1 · Q84

Q.If x=log⁡abc, y=log⁡bca, z=log⁡cabx=\log_a bc,\ y=\log_b ca,\ z=\log_c ab, then prove that 11+x+11+y+11+z=1\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}=1.

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x=log⁡a(bc), y=log⁡b(ca), z=log⁡c(ab)x=\log_a(bc),\ y=\log_b(ca),\ z=\log_c(ab).

1+x=log⁡aa+log⁡a(bc)=log⁡a(a⋅bc)=log⁡a(abc)1+x=\log_aa+\log_a(bc)=\log_a(a\cdot bc)=\log_a(abc).

Similarly 1+y=log⁡b(abc)1+y=\log_b(abc) and 1+z=log⁡c(abc)1+z=\log_c(abc).

Using the reciprocal change-of-base identity 1log⁡NM=log⁡MN\dfrac{1}{\log_N M}=\log_M N:

11+x=1log⁡a(abc)=log⁡abca\dfrac{1}{1+x}=\dfrac{1}{\log_a(abc)}=\log_{abc}a, and likewise 11+y=log⁡abcb\dfrac{1}{1+y}=\log_{abc}b, 11+z=log⁡abcc\dfrac{1}{1+z}=\log_{abc}c. …

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