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Question Number 8956 by j.masanja06@gmail.com last updated on 07/Nov/16
prove that;  log_(ab) x=((log_a x−log_b x)/(log_a x+log_b x))
$$\mathrm{prove}\:\mathrm{that}; \\ $$$$\mathrm{log}_{\mathrm{ab}} \mathrm{x}=\frac{\mathrm{log}_{\mathrm{a}} \mathrm{x}−\mathrm{log}_{\mathrm{b}} \mathrm{x}}{\mathrm{log}_{\mathrm{a}} \mathrm{x}+\mathrm{log}_{\mathrm{b}} \mathrm{x}} \\ $$
Commented by sou1618 last updated on 07/Nov/16
log_p q=((log_r q)/(log_r p))  ((log_a x−log_b x)/(log_a x+log_b x))=((log_a x−(((log_a x)/(log_a b))))/(log_a x+(((log_a x)/(log_a b)))))         =((  (((log_a x)/(log_a b)))  )/(  (((log_a x)/(log_a b)))  ))×((log_a b−1)/(log_a b+1))         =((log_a b−log_a a)/(log_a b+log_a a))         =((log_a (b/a))/(log_a ab))         =log_(ab) (b/a) (=constant)  ((log_a x−log_b x)/(log_a x+log_b x))≠log_(ab) x  // // // // // // // //  if  ((log_a x−log_b x)/(log_a x+log_b x))=log_(ab) x,  x=(b/a).
$${log}_{{p}} {q}=\frac{{log}_{{r}} {q}}{{log}_{{r}} {p}} \\ $$$$\frac{{log}_{{a}} {x}−{log}_{{b}} {x}}{{log}_{{a}} {x}+{log}_{{b}} {x}}=\frac{{log}_{{a}} {x}−\left(\frac{{log}_{{a}} {x}}{{log}_{{a}} {b}}\right)}{{log}_{{a}} {x}+\left(\frac{{log}_{{a}} {x}}{{log}_{{a}} {b}}\right)} \\ $$$$\:\:\:\:\:\:\:=\frac{\:\:\left(\frac{{log}_{{a}} {x}}{{log}_{{a}} {b}}\right)\:\:}{\:\:\left(\frac{{log}_{{a}} {x}}{{log}_{{a}} {b}}\right)\:\:}×\frac{{log}_{{a}} {b}−\mathrm{1}}{{log}_{{a}} {b}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\frac{{log}_{{a}} {b}−{log}_{{a}} {a}}{{log}_{{a}} {b}+{log}_{{a}} {a}} \\ $$$$\:\:\:\:\:\:\:=\frac{{log}_{{a}} \frac{{b}}{{a}}}{{log}_{{a}} {ab}} \\ $$$$\:\:\:\:\:\:\:={log}_{{ab}} \frac{{b}}{{a}}\:\left(={constant}\right) \\ $$$$\frac{{log}_{{a}} {x}−{log}_{{b}} {x}}{{log}_{{a}} {x}+{log}_{{b}} {x}}\neq{log}_{{ab}} {x} \\ $$$$//\://\://\://\://\://\://\:// \\ $$$${if} \\ $$$$\frac{{log}_{{a}} {x}−{log}_{{b}} {x}}{{log}_{{a}} {x}+{log}_{{b}} {x}}={log}_{{ab}} {x}, \\ $$$${x}=\frac{{b}}{{a}}. \\ $$$$ \\ $$

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