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log ​a ​ n log b n+log b n log c n+log a n log c n=(log a ​n log b n log c ​n)/log ​abc ​n. Prove

log​a​n logbn+logbn logcn+logan logcn=(loga​n logbn logc​n)/log​abc​n. Prove

Grade:10

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
7 years ago
we know thatlog_xy = log_{10}y/log_{10}x
we will us that here
LHS = \frac{log^2(n)}{log(a)log(b)} + \frac{log^2(n)}{log(b)log(c)} + \frac{log^2(n)}{log(a)log(c)}
LHS = \frac{log^2(n) \times (log(a) + log(b) + log(c))}{log(a)log(b)log(c)} = \frac{log^3(n) \times (log(a) + log(b) + log(c))}{log(a)log(b)log(c)log(n)}LHS = \frac{log_a(n) log_b(n) log_c(n) \times log(abc)}{log(n)} = \frac{log_a(n) log_b(n) log_c(n) }{log_{abc}(n)}= RHS
hence proved

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