Question Number 113800 by Aina Samuel Temidayo last updated on 15/Sep/20 $$\mathrm{log}\:\left(\mathrm{ab}\right)−\mathrm{log}\mid\mathrm{b}\mid\:= \\ $$$$ \\ $$$$\mathrm{A}.\:\mathrm{log}\left(\mathrm{a}\right)\:\mathrm{B}.\:\mathrm{log}\:\mid\mathrm{a}\mid\:\mathrm{C}.\:−\mathrm{log}\left(\mathrm{a}\right)\:\mathrm{D}. \\ $$$$\mathrm{None}\:\mathrm{of}\:\mathrm{these}. \\ $$ Answered by MJS_new last updated…
Question Number 113803 by Aina Samuel Temidayo last updated on 15/Sep/20 $$\mathrm{If}\:\mathrm{a}=\mathrm{log}_{\mathrm{24}} \mathrm{12},\:\mathrm{b}=\mathrm{log}_{\mathrm{36}} \mathrm{24}\:\mathrm{and} \\ $$$$\mathrm{c}=\mathrm{log}_{\mathrm{48}} \mathrm{36},\:\mathrm{then}\:\mathrm{1}+\mathrm{abc}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$ \\ $$$$\left(\mathrm{A}\right)\:\mathrm{2ab}\:\left(\mathrm{B}\right)\:\mathrm{2ac}\:\left(\mathrm{C}\right)\:\mathrm{2bc}\:\left(\mathrm{D}\right)\:\mathrm{0} \\ $$ Answered by…
Question Number 48202 by F_Nongue last updated on 20/Nov/18 $${how}\:{can}\:{I}\:{get}\:{the}\:{x}? \\ $$$$\left.{a}\right)\mathrm{3}^{{x}} +\mathrm{4}^{{x}} =\mathrm{5}^{{x}} \\ $$$$\left.{b}\right)\mathrm{7}^{\mathrm{6}−{x}} ={x}+\mathrm{2} \\ $$$$\left.{c}\right)\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{2}^{{x}} \\ $$$$\left.{d}\right)\mathrm{3}^{{x}−\mathrm{2}} =\frac{\mathrm{9}}{{x}} \\…
Question Number 113426 by bemath last updated on 13/Sep/20 Answered by bemath last updated on 13/Sep/20 $$\left(\mathrm{1}\right)\:\mathrm{3}^{\mathrm{3}{b}} \:=\:\mathrm{5}^{\mathrm{4}{a}} \:\rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}^{\mathrm{3}{b}} \right)=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{4}{a}} \right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{2}{b}+\mathrm{2}}…
Question Number 112808 by Aina Samuel Temidayo last updated on 09/Sep/20 Answered by 1549442205PVT last updated on 10/Sep/20 $$\mathrm{We}\:\mathrm{have}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{10}+\mathrm{2x}\right)=\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}−\mathrm{4}\right)+\mathrm{2} \\ $$$$\Leftrightarrow\left[\mathrm{log}_{\mathrm{2}} \left(\mathrm{10}+\mathrm{2x}\right)−\mathrm{2}\right]−\mathrm{log}_{\mathrm{10}} \left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0}\left(\mathrm{1}\right)…
Question Number 112773 by 675480065 last updated on 09/Sep/20 $$\int\frac{\:\sqrt{{x}}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\ $$$${Please}\:{help} \\ $$ Answered by MJS_new last updated on 09/Sep/20 $$\int\frac{\sqrt{{x}}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx}= \\…
Question Number 177360 by jlewis last updated on 04/Oct/22 $$\mathrm{1}/\mathrm{2}\:{log}_{\mathrm{4}} \mathrm{36}\:×{log}_{\mathrm{6}} \mathrm{64} \\ $$ Answered by TheHoneyCat last updated on 08/Oct/22 $$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{4}} \left(\mathrm{4}×\mathrm{9}\right)×\mathrm{log}_{\mathrm{6}} \left(\mathrm{6}×\mathrm{9}\right) \\…
Question Number 177361 by jlewis last updated on 04/Oct/22 $$\mathrm{4}\left(\mathrm{16}^{\left({x}+\mathrm{4}\right)} \:×\:\mathrm{5}.\mathrm{2}^{\mathrm{2}{x}} =\mathrm{13}\right. \\ $$ Answered by a.lgnaoui last updated on 04/Oct/22 $$\:\mathrm{2}^{\mathrm{2}} \left(\mathrm{2}^{\mathrm{4}} \right)^{\left({x}+\mathrm{4}\right)} ×\mathrm{5}.\mathrm{2}^{\mathrm{2}{x}}…
Question Number 111721 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{If}\:\mathrm{b}>\mathrm{1},\mathrm{x}>\mathrm{0}\:\mathrm{and}\:\left(\mathrm{2x}\right)^{\mathrm{log}_{\mathrm{b}} \mathrm{2}} −\left(\mathrm{3x}\right)^{\mathrm{log}_{\mathrm{b}} \mathrm{3}} =\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{x}\:\mathrm{is} \\ $$ Answered by Her_Majesty last updated…
Question Number 177113 by LOSER last updated on 01/Oct/22 $${X}\:{and}\:{Y}\:{are}\:{playing}\:{a}\:{game}.\: \\ $$$${Initially}\:{there}\:{are}\:{three}\:{bundles}\:{of}\: \\ $$$${matches},\:{consisting}\:{of}\:\mathrm{2021},\:\mathrm{2022}\: \\ $$$${and}\:\mathrm{2023}\:{pieces}.\:{Each}\:{player}\:{in}\:{his}\: \\ $$$${turn}\:{chooses}\:{a}\:{bundle}\:{B}\:{and}\:{removes}\: \\ $$$${a}\:{positive}\:{number}\:{of}\:{the}\:{matches}\:{of}\:{B}\: \\ $$$${such}\:{that}\:{the}\:{number}\:{of}\:{pieces}\:{of}\: \\ $$$${bundles}\:{still}\:{form}\:{an}\:{arithmetic}\: \\…