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Algebra> log a n log b n+log b n log c n+log a ...

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log_an log_bn+log_bn log_cn+log_an log_cn=(log_an log_bn log_cn)/log_abcn. Prove

krishanu , 9 Years ago

Grade 10

1 Answers

Riddhish Bhalodia

Last Activity: 9 Years ago

we know that $log_xy = log_{10}y/log_{10}x$
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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