Question Number 10186 by prakash jain last updated on 29/Jan/17 $$\mathrm{Prove} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} =\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{3}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}\centerdot\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{3}} }+..\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10064 by PradipGos. last updated on 22/Jan/17 $$\mathrm{if}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{3ab}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{log}\frac{\mathrm{a}+\mathrm{b}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}{a}+\mathrm{log}{b}\right) \\ $$$$ \\ $$ Answered by ridwan balatif last updated on…
Question Number 140730 by liberty last updated on 12/May/21 $$\mathrm{If}\:\mathrm{equation}\:\mathrm{2log}\:\left(\mathrm{x}+\mathrm{3}\right)=\mathrm{log}\:\mathrm{ax} \\ $$$$\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}.\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{a}. \\ $$ Answered by mr W last updated on 12/May/21 $${a}\neq\mathrm{0}…
Question Number 140680 by bemath last updated on 11/May/21 $$\:\mathrm{log}\:_{\left(\mathrm{x}+\mathrm{2}\right)} \left(\mathrm{7x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} \right)−\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}\right)\geqslant\:\mathrm{log}\:_{\sqrt{\mathrm{x}+\mathrm{2}}\:} \left(\sqrt{\mathrm{5}−\mathrm{x}}\:\right) \\ $$ Answered by liberty last updated on 11/May/21…
Question Number 9516 by Joel575 last updated on 12/Dec/16 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\lfloor\mathrm{log}_{\mathrm{2}} \:{k}\rfloor\:=\:\mathrm{2018} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{n}\:? \\ $$ Commented by sou1618 last updated on 12/Dec/16 $$\ast{K}=\lfloor{log}_{\mathrm{2}}…
Question Number 74945 by chess1 last updated on 04/Dec/19 Commented by chess1 last updated on 04/Dec/19 $$\mathrm{solve}\:\mathrm{equation} \\ $$ Answered by MJS last updated on…
Question Number 9334 by j.masanja06@gmail.com last updated on 30/Nov/16 $$\mathrm{express}\:\mathrm{the}\:\mathrm{following}\:\mathrm{in}\:\mathrm{form}\:\mathrm{of}\:\mathrm{log}\:\mathrm{a}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \\ $$$$\left(\mathrm{i}\right)\:\mathrm{4log}_{\mathrm{a}} \mathrm{x}−\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}^{\mathrm{3}} \right) \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{log}_{\mathrm{a}} \mathrm{x}\:+\:\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{x}+\mathrm{3}\right) \\ $$ Answered by mrW…
Question Number 9333 by j.masanja06@gmail.com last updated on 30/Nov/16 $$\mathrm{given}\:\mathrm{that}\:\mathrm{log}\:\mathrm{2}=\mathrm{0}.\mathrm{3}.\mathrm{show}\:\mathrm{that}\:\mathrm{log}\:\mathrm{5}=\mathrm{0}.\mathrm{7} \\ $$ Answered by mrW last updated on 30/Nov/16 $$\mathrm{2}×\mathrm{5}=\mathrm{10} \\ $$$$\mathrm{log}\:\left(\mathrm{2}×\mathrm{5}\right)=\mathrm{log}\:\mathrm{10}=\mathrm{1} \\ $$$$\mathrm{log}\:\mathrm{2}+\mathrm{log}\:\mathrm{5}=\mathrm{1} \\…
Question Number 74753 by necxxx last updated on 30/Nov/19 $${evaluate}\:\mathrm{5}^{\sqrt{\mathrm{log}\:\mathrm{7}_{\mathrm{5}} }} \:−\:\mathrm{7}^{\sqrt{\mathrm{log}\:\mathrm{5}_{\mathrm{7}} }} \\ $$ Commented by mr W last updated on 30/Nov/19 $${what}\:{is}\:\mathrm{log}\:\mathrm{7}_{\mathrm{5}} ?\:{the}\:{same}\:{as}\:\mathrm{log}_{\mathrm{5}}…
Question Number 74589 by Aditya789 last updated on 27/Nov/19 Answered by MJS last updated on 27/Nov/19 $${a}\left({b}−{c}\right){x}^{\mathrm{2}} +{b}\left({c}−{a}\right){xy}+{c}\left({a}−{b}\right){y}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{{b}\left({c}−{a}\right){y}}{{a}\left({b}−{c}\right)}{x}+\frac{{c}\left({a}−{b}\right){y}^{\mathrm{2}} }{{a}\left({b}−{c}\right)}=\mathrm{0} \\ $$$${x}={t}−\frac{{b}\left({c}−{a}\right){y}}{\mathrm{2}{a}\left({b}−{c}\right)}…