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Category: Algebra

Question-197419

Question Number 197419 by universe last updated on 16/Sep/23 Answered by cortano12 last updated on 17/Sep/23 $$\:\:\:\:\begin{cases}{\left(\mathrm{2a}\right)^{\mathrm{ln}\:\mathrm{a}} \:=\:\left(\mathrm{bc}\right)^{\mathrm{ln}\:\mathrm{b}} }\\{\mathrm{b}^{\mathrm{ln}\:\mathrm{2}} \:=\:\mathrm{a}^{\mathrm{ln}\:\mathrm{c}} \:}\end{cases} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{ln}\:\left(\mathrm{2a}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)\:=\:\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{bc}\right)}\\{\mathrm{ln}\:\left(\mathrm{b}\right).\:\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left(\mathrm{c}\right).\:\mathrm{ln}\:\left(\mathrm{a}\right)}\end{cases} \\ $$$$\:\:\Rightarrow\mathrm{ln}\:\mathrm{a}\:\left\{\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{a}\right\}=\:\mathrm{ln}\:\mathrm{b}\:\left\{\mathrm{ln}\:\mathrm{b}+\mathrm{ln}\:\mathrm{c}\:\right\}…

a-b-c-R-a-2b-3ac-3ac-a-4b-bc-b-Find-2b-a-3a-b-

Question Number 197248 by hardmath last updated on 11/Sep/23 $$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R} \\ $$$$\frac{\mathrm{a}\:+\:\mathrm{2b}\:−\:\mathrm{3ac}}{\mathrm{3ac}}\:\:=\:\:\frac{\mathrm{a}\:+\:\mathrm{4b}\:−\:\mathrm{bc}}{\mathrm{b}} \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{2b}}{\mathrm{a}}\:−\:\frac{\mathrm{3a}}{\mathrm{b}} \\ $$ Commented by Frix last updated on 11/Sep/23 $$−\frac{\left({c}+\mathrm{1}\right)\left(\mathrm{3}{c}^{\mathrm{2}} −\mathrm{15}{c}+\mathrm{1}\right)}{\mathrm{2}{c}}\pm\frac{\left({c}−\mathrm{1}\right)\sqrt{\mathrm{9}{c}^{\mathrm{4}}…

Question-197064

Question Number 197064 by Amidip last updated on 07/Sep/23 Answered by witcher3 last updated on 07/Sep/23 $$\mathrm{x}\sqrt{\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }+\mathrm{y}\left(\mathrm{1}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\mathrm{a} \\ $$$$\mathrm{a}^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}}…