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