Question Number 56418 by gunawan last updated on 16/Mar/19
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:{a}^{\mathrm{log}_{{b}} {x}} \mathrm{where}\:{a}=\mathrm{0}.\mathrm{2},\:{b}=\sqrt{\mathrm{5},} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\:+\:…\:\mathrm{to}\:\infty\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
$${x}=\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}.\mathrm{5} \\ $$$${t}={log}_{{b}} {x}\rightarrow{b}^{{t}} ={x}\rightarrow\left(\mathrm{5}\right)^{\frac{{t}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}.\mathrm{5} \\ $$$$\frac{{t}}{\mathrm{2}}={log}_{\mathrm{5}} \left(\mathrm{0}.\mathrm{5}\right)\rightarrow{t}=\mathrm{2}{log}_{\mathrm{5}} \left(\mathrm{0}.\mathrm{5}\right) \\ $$$${so}\:{a}^{{log}_{{b}} {x}} \rightarrow\mathrm{0}.\mathrm{2}^{{log}_{\mathrm{5}} \left(\mathrm{0}.\mathrm{25}\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$