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Category: Logarithms

The-result-log-xy-log-x-log-y-is-not-always-true-Give-a-pair-of-values-of-x-and-y-such-that-the-result-will-not-hold-

Question Number 5569 by Rasheed Soomro last updated on 20/May/16 $$\mathrm{The}\:\mathrm{result}\:“\mathrm{log}\:\mathrm{xy}=\mathrm{log}\:\mathrm{x}+\mathrm{log}\:\mathrm{y}''\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{always}\:\mathrm{true}!\:\mathrm{Give}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of} \\ $$$$\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{result}\:\mathrm{will}\:\mathrm{not}\:\mathrm{hold}. \\ $$ Commented by Yozzii last updated on 20/May/16 $$\left({x},{y}\right)=\left(−\mathrm{1},\mathrm{2}\right)…

ln-e-ln-e-ln-e-

Question Number 71016 by mr W last updated on 10/Oct/19 $$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)=? \\ $$ Answered by mr W last updated on 11/Oct/19 $$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+…\right)\right)\right)={x} \\ $$$$\boldsymbol{\mathrm{ln}}\:\left(\boldsymbol{{e}}+{x}\right)={x} \\…

log-x-2-10-3x-lt-2-

Question Number 136256 by EDWIN88 last updated on 20/Mar/21 $$\mathrm{log}\:_{\left(\mathrm{x}−\mathrm{2}\right)} \left(\mathrm{10}−\mathrm{3x}\right)\:<\:\mathrm{2} \\ $$ Answered by liberty last updated on 20/Mar/21 $$\left(\mathrm{1}\right)\:\mathrm{10}−\mathrm{3}{x}\:>\:\mathrm{0}\:;\:\mathrm{3}{x}−\mathrm{10}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{x}\:<\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{log}\:_{\left({x}−\mathrm{2}\right)}…

log-x-y-y-x-

Question Number 5039 by Rasheed Soomro last updated on 05/Apr/16 $$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=? \\ $$ Answered by LMTV last updated on 05/Apr/16 $$\left(\frac{{x}}{{y}}\right)^{?} =\frac{{y}}{{x}} \\ $$$$?=−\mathrm{1}…