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Question Number 17779 by Mr easymsn last updated on 10/Jul/17
show that {log_a ab}{log_b ab}=logab_a +logab_b
$${show}\:{that}\:\left\{{lo}\underset{{a}} {{g}ab}\right\}\left\{{lo}\underset{{b}} {{g}ab}\right\}={loga}\underset{{a}} {{b}}+{loga}\underset{{b}} {{b}} \\ $$
Answered by Tinkutara last updated on 11/Jul/17
(log_a  ab)(log_b  ab)  = (1 + log_a  b)(1 + log_b  a)  = 1 + log_b  a + log_a  b + 1  = log_b  b + log_b  a + log_a  b + log_a  a  = log_a  ab + log_b  ab
$$\left(\mathrm{log}_{{a}} \:{ab}\right)\left(\mathrm{log}_{{b}} \:{ab}\right) \\ $$$$=\:\left(\mathrm{1}\:+\:\mathrm{log}_{{a}} \:{b}\right)\left(\mathrm{1}\:+\:\mathrm{log}_{{b}} \:{a}\right) \\ $$$$=\:\mathrm{1}\:+\:\mathrm{log}_{{b}} \:{a}\:+\:\mathrm{log}_{{a}} \:{b}\:+\:\mathrm{1} \\ $$$$=\:\mathrm{log}_{{b}} \:{b}\:+\:\mathrm{log}_{{b}} \:{a}\:+\:\mathrm{log}_{{a}} \:{b}\:+\:\mathrm{log}_{{a}} \:{a} \\ $$$$=\:\boldsymbol{\mathrm{log}}_{\boldsymbol{{a}}} \:\boldsymbol{{ab}}\:+\:\boldsymbol{\mathrm{log}}_{\boldsymbol{{b}}} \:\boldsymbol{{ab}} \\ $$

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