Question Number 210607 by peter frank last updated on 13/Aug/24
Answered by A5T last updated on 14/Aug/24
$$\mathrm{75}\left(\mathrm{5}^{\mathrm{3}{log}_{\mathrm{2}} {x}} \right)=\mathrm{75}\left(\mathrm{2}^{{log}_{\mathrm{2}} \mathrm{5}} \right)^{{log}_{\mathrm{2}} {x}×\mathrm{3}} =\mathrm{75}\left(\mathrm{2}^{{log}_{\mathrm{2}} {x}} \right)^{\mathrm{3}{log}_{\mathrm{2}} \mathrm{5}} \\ $$$$=\mathrm{75}\left({x}\right)^{\mathrm{3}{log}_{\mathrm{2}} \mathrm{5}} =\mathrm{4}{x} \\ $$$$\Rightarrow{x}^{{log}_{\mathrm{2}} \left(\frac{\mathrm{125}}{\mathrm{2}}\right)} =\frac{\mathrm{4}}{\mathrm{75}}\Rightarrow{x}=\left(\frac{\mathrm{4}}{\mathrm{75}}\right)^{\frac{\mathrm{1}}{{log}_{\mathrm{2}} \left(\mathrm{125}\right)−\mathrm{1}}} \approx\mathrm{0}.\mathrm{6118} \\ $$
Commented by peter frank last updated on 14/Aug/24
Commented by peter frank last updated on 14/Aug/24
$$\mathrm{where}\:\mathrm{did}\:\mathrm{that}\:\mathrm{2}\:\mathrm{come}\:\mathrm{from}? \\ $$
Commented by A5T last updated on 14/Aug/24
$${b}={a}^{{log}_{{a}} {b}} \Rightarrow\mathrm{5}=\mathrm{2}^{{log}_{\mathrm{2}} \mathrm{5}} \\ $$
Answered by Spillover last updated on 14/Aug/24
Answered by Spillover last updated on 14/Aug/24
