Question Number 146677 by mathdanisur last updated on 14/Jul/21
$${if}\:\:\:{f}\left({x}\right)\:=\:\mathrm{3}^{\boldsymbol{{x}}+\mathrm{1}} \:\:\:{find}\:\:\:\frac{{f}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)}{{f}\left({x}\:+\:\mathrm{1}\right)}\:=\:? \\ $$
Answered by hknkrc46 last updated on 14/Jul/21
$$\blacktriangleright\:\boldsymbol{{f}}\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:=\:\mathrm{3}^{\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:+\:\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{2}} \\ $$$$\blacktriangleright\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:=\:\mathrm{3}^{\left(\boldsymbol{{x}}\:+\:\mathrm{1}\right)\:+\:\mathrm{1}} \:=\:\mathrm{3}^{\boldsymbol{{x}}\:+\:\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\boldsymbol{{f}}\left(\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1}\right)}{\boldsymbol{{f}}\left(\boldsymbol{{x}}\:+\:\mathrm{1}\right)}\:=\:\frac{\mathrm{3}^{\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{2}} }{\mathrm{3}^{\boldsymbol{{x}}\:+\:\mathrm{2}} }\:=\:\mathrm{3}^{\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{2}\:−\:\left(\boldsymbol{{x}}\:+\:\mathrm{2}\right)} \\ $$$$=\:\mathrm{3}^{\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{2}\:−\:\boldsymbol{{x}}\:−\:\mathrm{2}} \:=\:\mathrm{3}^{\boldsymbol{{x}}} \:=\:\frac{\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 15/Jul/21
$${thankyou}\:{Ser}\:{cool} \\ $$