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Category: Logarithms

Question-171031

Question Number 171031 by cortano1 last updated on 06/Jun/22 Answered by a.lgnaoui last updated on 07/Jun/22 $$={Log}\left(\frac{{a}}{\mathrm{1}−{b}}\right)+{Log}\left(\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}−{b}}×\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\right)={Log}\left(\frac{{a}}{\mathrm{1}+{b}}\right) \\ $$$$={Log}\left(\frac{{a}−{b}+\mathrm{1}}{{a}+{b}+\mathrm{1}}\right)^{\mathrm{3}} =\mathrm{3}{Log}\left(\frac{{a}+\mathrm{1}−{b}}{{a}+\mathrm{1}+{b}}\right)=\mathrm{3}{Log}\left(\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} }{\left({a}+\mathrm{1}+{b}\right)^{\mathrm{2}} }\right) \\ $$$$=\mathrm{3}{Log}\left[\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}}…

log-5-x-3-log-6-x-14-

Question Number 170958 by cortano1 last updated on 04/Jun/22 $$\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{5}} \left({x}+\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{6}} \left({x}+\mathrm{14}\right) \\ $$ Answered by floor(10²Eta[1]) last updated on 05/Jun/22 $$\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}+\mathrm{3}\right)=\mathrm{y}=\mathrm{log}_{\mathrm{6}} \left(\mathrm{x}+\mathrm{14}\right) \\…

Question-170843

Question Number 170843 by cortano1 last updated on 01/Jun/22 Answered by floor(10²Eta[1]) last updated on 01/Jun/22 $$\frac{\mathrm{log}\left(\mathrm{x}−\mathrm{40}\right)}{\mathrm{log}\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}}=\mathrm{1}\Leftrightarrow\mathrm{log}\left(\mathrm{x}−\mathrm{40}\right)=\mathrm{log}\left(\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}\right) \\ $$$$\Leftrightarrow\mathrm{x}−\mathrm{40}=\sqrt{\mathrm{x}+\mathrm{1}}+\mathrm{1}\Leftrightarrow\sqrt{\mathrm{x}+\mathrm{1}}=\mathrm{x}−\mathrm{41} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{1}=\left(\mathrm{x}−\mathrm{41}\right)^{\mathrm{2}} \wedge\mathrm{x}−\mathrm{41}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{83x}+\mathrm{41}^{\mathrm{2}}…

a-b-c-are-real-numbers-1-lt-b-lt-c-2-lt-a-10-log-a-b-2log-b-c-5log-c-a-12-prove-that-2log-a-c-5log-c-b-10log-b-a-21-

Question Number 105269 by malwaan last updated on 27/Jul/20 $${a};{b};{c}\:{are}\:{real}\:{numbers} \\ $$$$\mathrm{1}<{b}<{c}^{\mathrm{2}} <{a}^{\mathrm{10}} \\ $$$${log}_{{a}} {b}+\mathrm{2}{log}_{{b}} {c}+\mathrm{5}{log}_{{c}} {a}=\mathrm{12} \\ $$$${prove}\:{that} \\ $$$$\mathrm{2}{log}_{{a}} {c}+\mathrm{5}{log}_{{c}} {b}+\mathrm{10}{log}_{{b}} {a}\geqslant\mathrm{21}…

Question-170776

Question Number 170776 by solomonwells last updated on 30/May/22 Commented by mr W last updated on 31/May/22 $${i}\:{think}\:{there}\:{is}\:{no}\:“{nice}''\:{solution}. \\ $$$$\:{you}\:{can}\:{only}\:{approximate}.\:{x}\approx\mathrm{517}.\mathrm{9788}. \\ $$$${this}\:{wouldn}'{t}\:{change}\:{even}\:{when}\:{you}\: \\ $$$${repeat}\:{the}\:{same}\:{question}\:{a}\:{pair}\: \\…