Question Number 41495 by Dawajan Nikmal last updated on 08/Aug/18
$$\mathrm{The}\:\mathrm{equation}\:\pi^{{x}} =−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{9}\:\mathrm{has} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
$${no}\:{solution} \\ $$
Commented by math khazana by abdo last updated on 10/Aug/18
$${first}\:{what}\:{is}\:{the}\:{question}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
$${to}\:{find}\:{the}\:{value}\:{of}\:{x} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
Answered by MrW3 last updated on 09/Aug/18
$${RHS}=−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{9}=−\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{4}}\right)+\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{9}=−\mathrm{2}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{2}}\leqslant−\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${LHS}=\pi^{{x}} >\mathrm{0} \\ $$$${since}\:{LHS}>\mathrm{0}\:{and}\:{RHS}\leqslant−\frac{\mathrm{9}}{\mathrm{2}}, \\ $$$${there}\:{is}\:{no}\:{solution}\:{for}\:{the}\:{eqn}. \\ $$