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LogarithmsQuestion and Answers: Page 8
Question Number 143812 Answers: 1 Comments: 0
$$\:\mathrm{log}\:_{\mathrm{a}} \left(\mathrm{ax}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{ax}\right)=\mathrm{log}\:_{\mathrm{a}^{\mathrm{2}} } \left(\frac{\mathrm{1}}{\mathrm{a}}\right) \\ $$$$\:\mathrm{a}>\mathrm{0}\:,\:\mathrm{a}\neq\mathrm{1}\:.\:\mathrm{So}\:\mathrm{x}\:=\:? \\ $$
Question Number 143811 Answers: 1 Comments: 0
$$\:\begin{cases}{\mathrm{5}\left(\mathrm{log}\:_{\mathrm{y}} \left(\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{y}\right)\right)=\mathrm{26}}\\{\:\mathrm{xy}\:=\:\mathrm{64}}\end{cases}\mathrm{then} \\ $$$$\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{xy}\:=? \\ $$
Question Number 143475 Answers: 1 Comments: 1
$$\:{evaluate}; \\ $$$$\frac{\left(\sqrt{\mathrm{7}}\right)^{\mathrm{log64}} −\left(\mathrm{3}\right)^{\mathrm{log}_{\mathrm{24}} \mathrm{8}} }{\left(\mathrm{log}\:_{\mathrm{2}} \mathrm{8}−\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{64}\right)\left(\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{64}}\right)}\right)} \\ $$
Question Number 145524 Answers: 1 Comments: 0
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{argument}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{below} \\ $$$$\left(\mathrm{i}\right)\:{z}\:=\:\mathrm{1}+{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$$$\left(\mathrm{ii}\right)\:{z}\:=\:\mathrm{1}\:−{e}^{\frac{\pi}{\mathrm{6}}{i}} \\ $$
Question Number 142755 Answers: 1 Comments: 2
$$\mathrm{log}_{\mathrm{3}} \mathrm{x}^{\mathrm{3}} +\mathrm{log}_{\mathrm{2}} \mathrm{x}^{\mathrm{2}} =\frac{\mathrm{2lg6}}{\mathrm{lg2}}+\mathrm{1}\:\:\mathrm{find}\:\:\mathrm{x} \\ $$
Question Number 141365 Answers: 1 Comments: 0
$$\:\mathrm{log}\:_{\frac{\mathrm{9}}{\mathrm{4}}} \left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{\mathrm{6}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\sqrt{...}}}}\right)\:=? \\ $$
Question Number 140730 Answers: 1 Comments: 0
$$\mathrm{If}\:\mathrm{equation}\:\mathrm{2log}\:\left(\mathrm{x}+\mathrm{3}\right)=\mathrm{log}\:\mathrm{ax} \\ $$$$\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}.\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{a}. \\ $$
Question Number 140680 Answers: 1 Comments: 0
$$\:\mathrm{log}\:_{\left(\mathrm{x}+\mathrm{2}\right)} \left(\mathrm{7x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} \right)−\mathrm{log}\:_{\frac{\mathrm{1}}{\mathrm{x}+\mathrm{2}}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}\right)\geqslant\:\mathrm{log}\:_{\sqrt{\mathrm{x}+\mathrm{2}}\:} \left(\sqrt{\mathrm{5}−\mathrm{x}}\:\right) \\ $$
Question Number 138091 Answers: 2 Comments: 0
$${If}\:\mathrm{3}^{\left(\mathrm{log}\:_{\mathrm{3}} \:\mathrm{7}\right)^{{x}} } \:=\:\mathrm{7}^{\left(\mathrm{log}\:_{\mathrm{7}} \:\mathrm{3}\right)^{{x}} } \:,\:{then}\:{the} \\ $$$${value}\:{of}\:{x}\:{will}\:{be}\:\ldots\: \\ $$
Question Number 137764 Answers: 1 Comments: 0
$$\mathrm{If}\:\mathrm{log}_{\mathrm{2}} \mathrm{3}={a}\:\mathrm{and}\:\mathrm{log}_{\mathrm{3}} \mathrm{7}={b},\:\mathrm{express} \\ $$$$\mathrm{log}_{\mathrm{42}} \mathrm{56}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b} \\ $$
Question Number 136256 Answers: 1 Comments: 0
$$\mathrm{log}\:_{\left(\mathrm{x}−\mathrm{2}\right)} \left(\mathrm{10}−\mathrm{3x}\right)\:<\:\mathrm{2} \\ $$
Question Number 136064 Answers: 1 Comments: 0
$$\:\:\:\:\:\mathrm{3}^{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}^{{x}} −\mathrm{1}\right)} \:=\:\mathrm{2}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}^{{x}} +\mathrm{1}\right)} +\:\mathrm{1} \\ $$
Question Number 134767 Answers: 0 Comments: 0
Question Number 134116 Answers: 1 Comments: 0
Question Number 133936 Answers: 2 Comments: 0
$$\mathrm{log}\:_{\mathrm{x}+\mathrm{8}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{4}\right)\:<\:\mathrm{2}.\mathrm{log}\:_{\left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{2}} } \left(\mid\mathrm{x}−\mathrm{4}\mid\right)\: \\ $$
Question Number 133615 Answers: 2 Comments: 0
$$\mathrm{If}\:\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{log}\:_{\mathrm{2}} \:\mathrm{x}\right)+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{log}\:_{\mathrm{4}} \:\mathrm{x}\right)=\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{log}\:_{\mathrm{5}} \:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}\:+\mathrm{5}}\:=\:? \\ $$
Question Number 131757 Answers: 0 Comments: 0
$${log}\mathrm{3}\sqrt{\mathrm{5}} \\ $$
Question Number 131661 Answers: 1 Comments: 0
$$\:\:\:\mathrm{log}\:_{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}} \left(\mathrm{1}+\mathrm{cos}\:\mathrm{x}\right)\:=\:\mathrm{2} \\ $$$$\:−\frac{\mathrm{2}\pi}{\mathrm{3}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{3}} \\ $$
Question Number 131457 Answers: 1 Comments: 0
$$\mathrm{Find}\:{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{3}^{{x}+\mathrm{1}} \:=\:\mathrm{4}^{{x}−\mathrm{1}} \\ $$
Question Number 131890 Answers: 1 Comments: 0
$${log}_{\mathrm{2}} {x}+{log}_{\mathrm{3}} {x}=\mathrm{1}\:\:\:\:\:{x}=? \\ $$
Question Number 130858 Answers: 0 Comments: 0
$$\mathrm{log}\:\left[\sqrt{\boldsymbol{{mx}}^{\mathrm{2}} +\mathrm{3}}\:+\boldsymbol{{x}}\right]=? \\ $$The following subject is a node subject with a value of m
Question Number 130553 Answers: 1 Comments: 0
$${log}\left({x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+....\right)=? \\ $$
Question Number 130244 Answers: 0 Comments: 3
Question Number 129451 Answers: 1 Comments: 0
$$\:\mathrm{log}\:_{\mathrm{12}} \left(\sqrt{\mathrm{x}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{x}}\:\right)\:=\:\mathrm{log}\:_{\mathrm{9}} \left(\sqrt{\mathrm{x}}\:\right)\: \\ $$$$\:\mathrm{x}\:=\:? \\ $$
Question Number 129088 Answers: 1 Comments: 0
Question Number 128829 Answers: 0 Comments: 0
$$\:\mathrm{If}\:\mathrm{x},\mathrm{y}\:\mathrm{and}\:\mathrm{x}\:\mathrm{in}\:\mathrm{HP}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{log}\:\left(\mathrm{x}+\mathrm{z}\right)\:+\mathrm{log}\:\left(\mathrm{x}+\mathrm{z}−\mathrm{y}\right)\:=\:\mathrm{2}\:\mathrm{log}\:\left(\mathrm{x}−\mathrm{z}\right) \\ $$
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