Question Number 25313 by ibraheem160 last updated on 08/Dec/17
$${show}\:{tbat}\:{log}_{{a}} ^{\left({a}^{\mathrm{2}} −{x}^{\left.\mathrm{2}\right)} \right.} =\mathrm{2}+{log}_{{a}} \left[\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right] \\ $$
Answered by jota+ last updated on 08/Dec/17
$$\mathrm{2}+\mathrm{log}_{{a}} \:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)= \\ $$$$\mathrm{2}+\mathrm{log}_{{a}} \:\left(\frac{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)= \\ $$$$\mathrm{2}+\mathrm{log}\:_{{a}} \left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)−\mathrm{log}_{{a}} \:{a}^{\mathrm{2}} = \\ $$$$\:\:\:\mathrm{log}\:_{{a}} \left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right). \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 08/Dec/17
$$\mathcal{N}{ice}\:\mathcal{S}{ir}! \\ $$