Question Number 35899 by imamu222 last updated on 25/May/18
$${Evaluate}\:{log}_{\sqrt{\mathrm{2}}} \mathrm{4}+{log}_{\mathrm{1}/\mathrm{2}} \mathrm{16}−{log}_{\mathrm{4}} \mathrm{32} \\ $$
Commented by Cheyboy last updated on 26/May/18
$${log}_{\sqrt{\mathrm{2}}} \mathrm{4}+{log}_{\mathrm{1}/\mathrm{2}} \mathrm{16}−{log}_{\mathrm{4}} \mathrm{32} \\ $$$${log}_{\mathrm{2}^{\mathrm{1}/\mathrm{2}} } \mathrm{4}+{log}_{\mathrm{2}^{−\mathrm{1}} } \mathrm{16}−{log}_{\mathrm{2}^{\mathrm{2}} } \mathrm{32} \\ $$$$\mathrm{2}{log}_{\mathrm{2}} \mathrm{4}−{log}_{\mathrm{2}} \mathrm{16}−\mathrm{1}/\mathrm{2}{log}_{\mathrm{2}} \mathrm{32} \\ $$$${log}_{\mathrm{2}} \mathrm{4}^{\mathrm{2}} −{log}_{\mathrm{2}} \mathrm{16}−{log}_{\mathrm{2}} \left(\sqrt{\mathrm{32}}\right) \\ $$$${log}_{\mathrm{2}} \mathrm{16}−{log}_{\mathrm{2}} \mathrm{16}−{log}_{\mathrm{2}} \left(\sqrt{\mathrm{32}}\right) \\ $$$${log}_{\mathrm{2}} \left(\mathrm{16}\boldsymbol{\div}\mathrm{16}\boldsymbol{\div}\sqrt{\mathrm{32}}\right) \\ $$$${log}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{32}}}\right)\Rightarrow{log}_{\mathrm{2}} \mathrm{2}^{−\mathrm{5}} ={x} \\ $$$$\mathrm{2}^{−\mathrm{5}/\mathrm{2}} =\mathrm{2}^{{x}} \\ $$$${x}=\:−\mathrm{5}/\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Commented by Cheyboy last updated on 26/May/18
$${ooh}\:{Thank}\:{you}\:{sir}\:{u}\:{are}\:{right}\: \\ $$$${i}\:{made}\:{mistake} \\ $$