Question Number 211099 by peter frank last updated on 27/Aug/24
Answered by A5T last updated on 27/Aug/24
$$\mathrm{2}^{{logx}} =\mathrm{2}^{\frac{{log}_{\mathrm{2}} {x}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{\frac{\mathrm{1}}{{log}_{\mathrm{2}} \mathrm{10}}} ={x}^{{log}_{\mathrm{10}} \mathrm{2}} =\frac{{x}}{\mathrm{25}} \\ $$$$\Rightarrow{x}^{\mathrm{1}−{log}\mathrm{2}} =\mathrm{25}\Rightarrow{x}=\mathrm{25}^{\frac{\mathrm{1}}{\mathrm{1}{og}\mathrm{5}}} =\mathrm{25}^{{log}_{\mathrm{5}} \mathrm{10}} =\left(\mathrm{5}^{{log}_{\mathrm{5}} \mathrm{10}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{10}^{\mathrm{2}} =\mathrm{100} \\ $$
Commented by peter frank last updated on 28/Aug/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mr W last updated on 27/Aug/24
$$\left(\mathrm{log}\:\mathrm{2}\right)\mathrm{log}\:{x}=\mathrm{log}\:{x}−\mathrm{log}\:\mathrm{25} \\ $$$$\left(\mathrm{1}−\mathrm{log}\:\mathrm{2}\right)\mathrm{log}\:{x}=\mathrm{log}\:\mathrm{25} \\ $$$$\mathrm{log}\:{x}=\frac{\mathrm{log}\:\mathrm{25}}{\mathrm{1}−\mathrm{log}\:\mathrm{2}}=\frac{\mathrm{2}\:\mathrm{log}\:\mathrm{5}}{\mathrm{log}\:\mathrm{5}}=\mathrm{2}\: \\ $$$$\Rightarrow{x}=\mathrm{10}^{\mathrm{2}} =\mathrm{100} \\ $$
Commented by peter frank last updated on 28/Aug/24
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by mm1342 last updated on 27/Aug/24
$${logx}×{log}\mathrm{2}={logx}−{log}\mathrm{25} \\ $$$$\Rightarrow{logx}\left(\mathrm{1}−{log}\mathrm{2}\right)={log}\mathrm{25} \\ $$$$\Rightarrow{logx}=\frac{{log}\mathrm{25}}{{log}\mathrm{5}}=\mathrm{2}\Rightarrow{x}=\mathrm{100}\:\checkmark \\ $$$$ \\ $$
Commented by peter frank last updated on 28/Aug/24
$$\mathrm{thank}\:\mathrm{you} \\ $$