Question Number 6726 by Tawakalitu. last updated on 16/Jul/16
$${Find}\:{the}\:{value}\:{of}\:{x} \\ $$$$ \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{16}^{{x}} \\ $$$$ \\ $$$${please}\:{help}\:{with}\:{workings}. \\ $$
Answered by sou1618 last updated on 17/Jul/16
$${x}^{\mathrm{2}} =\mathrm{16}^{{x}} \rightarrow\left(\ast\right) \\ $$$${x}^{\mathrm{2}} =\left(\mathrm{4}^{{x}} \right)^{\mathrm{2}} \\ $$$$\mid{x}\mid=\mathrm{4}^{{x}} \left(\because\mathrm{4}^{{x}} >\mathrm{0}\right) \\ $$$$\mathrm{set} \\ $$$${f}\left({x}\right)=\mathrm{4}^{{x}} −\mid{x}\mid \\ $$$$ \\ $$$$\mathrm{when}\:{x}>=\mathrm{0} \\ $$$$\:\:\:\:{f}\:'\left({x}\right)=\left({log}_{{e}} \mathrm{4}\right)×\mathrm{4}^{{x}} −\mathrm{1}>\mathrm{0} \\ $$$$\:\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\:\:\:\therefore{f}\left(\mathrm{0}\right)>\mathrm{0}\:\Rightarrow\left(\ast\right)\mathrm{has}\:\mathrm{no}\:\mathrm{solution} \\ $$$$ \\ $$$$\mathrm{when}\:{x}<\mathrm{0} \\ $$$$\:\:\:{f}\:'\left({x}\right)=\left({log}_{{e}} \mathrm{4}\right)×\mathrm{4}^{{x}} +\mathrm{1}>\mathrm{0} \\ $$$$\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)=−\infty \\ $$$$\:\:\Rightarrow\left(\ast\right)\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\:\:\:\:−{x}=\mathrm{4}^{{x}} \\ $$$$\:\:\:\left(−{x}\right)^{\frac{\mathrm{1}}{{x}}} =\mathrm{4} \\ $$$$\:\:\:\mathrm{4}=\:\mathrm{2}^{\mathrm{2}} \:=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{2}} \: \\ $$$$\:\:\begin{cases}{−{x}=\mathrm{1}/\mathrm{2}}\\{\frac{\mathrm{1}}{{x}}=−\mathrm{2}}\end{cases} \\ $$$$\:{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Tawakalitu. last updated on 17/Jul/16
$${Thanks}\:{so}\:{much}. \\ $$