Question Number 111374 by mathdave last updated on 03/Sep/20 | ||
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Commented by kaivan.ahmadi last updated on 03/Sep/20 | ||
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$$={log}_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{14}}−\sqrt{\mathrm{5}}\right)+{log}_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{14}}+\sqrt{\mathrm{5}}\right)= \\ $$$${log}_{\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{14}}−\sqrt{\mathrm{5}}\right)\left(\sqrt{\mathrm{14}}+\sqrt{\mathrm{5}}\right)={log}_{\sqrt{\mathrm{3}}} \left(\mathrm{14}−\mathrm{5}\right)= \\ $$$${log}_{\sqrt{\mathrm{3}}} \mathrm{9}={log}_{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} } \mathrm{3}^{\mathrm{2}} =\mathrm{2}×\mathrm{2}{log}_{\mathrm{3}} \mathrm{3}=\mathrm{4} \\ $$ | ||
Answered by mnjuly1970 last updated on 03/Sep/20 | ||
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$$\:{ans}:\mathrm{4}\:\: \\ $$ | ||