Question Number 213923 by luj last updated on 21/Nov/24 Answered by MATHEMATICSAM last updated on 21/Nov/24 $${x}^{\mathrm{32}} \:=\:\mathrm{2}^{{x}} \\ $$$${or}\:\mathrm{32ln}\mid{x}\mid\:=\:{x}\mathrm{ln2} \\ $$$$\mathrm{For}\:{x}\:>\:\mathrm{0} \\ $$$${or}\:\frac{\mathrm{ln}{x}}{{x}}\:=\:\frac{\mathrm{ln2}}{\mathrm{32}}\: \\…
Question Number 212645 by efronzo1 last updated on 20/Oct/24 $$\:\:\begin{cases}{\mathrm{x}=\mathrm{2}+\:\mathrm{log}\:_{\mathrm{2}} \mathrm{log}\:_{\mathrm{2}} \mathrm{y}}\\{\mathrm{y}=\mathrm{2}\:\mathrm{log}\:_{\mathrm{2}} \mathrm{z}\:}\\{\mathrm{z}=\mathrm{2}+\:\mathrm{log}\:_{\mathrm{2}} \:\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212515 by efronzo1 last updated on 16/Oct/24 $$\:\:\mathrm{The}\:\mathrm{numbers}\:\mathrm{of}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{natural}\: \\ $$$$\:\:\:\mathrm{numbers}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{with}\:\mathrm{x},\mathrm{y}\:\leqslant\:\mathrm{33}\:\mathrm{that}\: \\ $$$$\:\:\:\mathrm{satisfy}\:\mathrm{5}\:\mid\:\mathrm{3}^{\mathrm{x}^{\mathrm{y}−\mathrm{1}} } \:+\:\mathrm{2}^{\mathrm{y}^{\mathrm{2x}} } \:\mathrm{is}\:…\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\mathrm{295}\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{296}\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{297}\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{298}\:\:\:\left(\mathrm{E}\right)\:\mathrm{299} \\ $$ Commented by A5T…
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Question Number 211815 by alcohol last updated on 21/Sep/24 $${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:{roots}\:\alpha\:{and}\:\beta \\ $$$${and}\:\frac{\alpha}{\beta}=\frac{\lambda}{\mu}.\:{show}\:{that}\:\lambda\mu{b}^{\mathrm{2}} \:=\:{ac}\left(\lambda+\mu\right)^{\mathrm{2}} \\ $$ Answered by som(math1967) last updated on 22/Sep/24 $$\:{let}\:\alpha={k}\lambda\:,\:\beta={k}\mu \\…
Question Number 209991 by efronzo1 last updated on 28/Jul/24 $$\:\:\:\:\:\frac{\mathrm{10}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{6}\right)} .\:\mathrm{15}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)} }{\mathrm{6}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)} .\:\mathrm{5}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)} }\:=?\: \\ $$ Answered by Rasheed.Sindhi last updated…
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Question Number 208412 by efronzo1 last updated on 15/Jun/24 Answered by A5T last updated on 15/Jun/24 $$\mathrm{8}=\mathrm{24}^{\frac{\mathrm{1}}{{a}}} ,\mathrm{27}=\mathrm{24}^{\frac{\mathrm{1}}{{b}}} ,\mathrm{64}=\mathrm{24}^{\frac{\mathrm{1}}{{c}}} \\ $$$$\Rightarrow\mathrm{24}^{\mathrm{3}} =\mathrm{8}×\mathrm{27}×\mathrm{64}=\mathrm{24}^{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{{ab}+{bc}+{ca}}{{abc}}\right)} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}}{{abc}}=\mathrm{3}\Rightarrow\frac{\mathrm{2022}{abc}}{{ab}+{bc}+{ca}}=\mathrm{2022}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{674} \\…
Question Number 208384 by efronzo1 last updated on 14/Jun/24 $$\:\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by A5T last updated on 14/Jun/24 $${log}_{{abc}} \left({a}\right)+{log}_{{abc}} \left({b}\right)=\mathrm{2}+\mathrm{3}=\mathrm{5}\Rightarrow{log}_{{abc}} \left({ab}\right)=\mathrm{5} \\ $$$${log}_{{abc}}…
Question Number 208322 by alcohol last updated on 11/Jun/24 Answered by Berbere last updated on 11/Jun/24 $$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)^{{k}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\frac{\left(\mathrm{1}+{k}\right)^{{k}} }{{k}^{{k}} }=\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}}…