Question Number 223685 by fantastic last updated on 02/Aug/25
$$\mathrm{25}^{{x}} −\mathrm{8}.\mathrm{5}^{{x}} =−\mathrm{16} \\ $$
Answered by Rasheed.Sindhi last updated on 02/Aug/25
$$\left(\mathrm{5}^{{x}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{5}^{{x}} \right)=−\mathrm{16} \\ $$$${let}\:\mathrm{5}^{{x}} ={t} \\ $$$${t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{16}=\mathrm{0} \\ $$$$\left({t}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${t}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{5}^{{x}} =\mathrm{4} \\ $$$${x}\mathrm{log}\:\mathrm{5}=\mathrm{log}\:\mathrm{4} \\ $$$${x}=\frac{\mathrm{log}\:\mathrm{4}}{\mathrm{log}\:\mathrm{5}} \\ $$$${x}=\mathrm{log}_{\mathrm{5}} \mathrm{4}\: \\ $$
Commented by fantastic last updated on 02/Aug/25
$$\vee.\cap{i}\mathbb{C}\in! \\ $$$${Seeing}\:{you}\:{after}\:{a}\:{long}\:{time}! \\ $$
Commented by Rasheed.Sindhi last updated on 02/Aug/25
$${Sir},\:{I}\:{try}\:{only}\:{selective}\:{questions}\:{so}\:{you} \\ $$$${see}\:{me}\:{very}\:{rarely}\:{on}\:{this}\:{forum}. \\ $$$${Anyway}\:{thank}\:{you}\:{to}\:{miss}\:{me}. \\ $$
Commented by fantastic last updated on 02/Aug/25
$${Oh}!\: \\ $$$$\left.:\right) \\ $$