Question Number 167746 by mathlove last updated on 24/Mar/22
$${log}_{\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)=? \\ $$
Commented by MJS_new last updated on 24/Mar/22
$$\mathrm{log}_{{x}^{\mathrm{2}} +\mathrm{2}} \:\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)\:=\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)}=\frac{\mathrm{ln}\:{x}\:+\mathrm{ln}\:\left({x}+\mathrm{4}\right)}{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \\ $$$$\Rightarrow \\ $$$${x}>\mathrm{0} \\ $$
Answered by LEKOUMA last updated on 24/Mar/22
$$\mathrm{log}\:_{\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)=\frac{\mathrm{log}\:\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{log}\:\left({x}^{\mathrm{2}} +\mathrm{2}\right)} \\ $$
Commented by MJS_new last updated on 24/Mar/22
$$\mathrm{you}\:\mathrm{cannot}\:\mathrm{find}\:{x}\:\mathrm{because}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{equation} \\ $$