Question Number 116491 by bemath last updated on 04/Oct/20 $$\left(\mathrm{0}.\mathrm{16}\right)^{\mathrm{log}\:_{\mathrm{2}.\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+…\right)} \:=? \\ $$ Answered by bobhans last updated on 04/Oct/20 $$\mathrm{Only}\:\mathrm{applying}\:\mathrm{property}\:\mathrm{of}\:\mathrm{logarithm} \\…
Question Number 50657 by cesar.marval.larez@gmail.com last updated on 18/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116085 by Ar Brandon last updated on 30/Sep/20 $$\mathrm{6}+\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{2}}} \left\{\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\sqrt{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\sqrt{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\sqrt{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\centerdot\centerdot\centerdot}}}\right\}=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 30/Sep/20 $$\sqrt{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\sqrt{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}….}}=\mathrm{a} \\ $$$$\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\mathrm{a}=\mathrm{a}^{\mathrm{2}}…
Question Number 181579 by mathlove last updated on 27/Nov/22 $$\left(\sqrt[{\mathrm{3}}]{{x}}\right)^{−\mathrm{2}+{log}_{{x}} \mathrm{11}} =\mathrm{11} \\ $$$${x}=? \\ $$ Commented by Socracious last updated on 27/Nov/22 $$\boldsymbol{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{11}} \\…
Question Number 115859 by bemath last updated on 29/Sep/20 $${Determine},\:{in}\:{simplest}\:{form}\:{the} \\ $$$${smallest}\:{of}\:{the}\:{three}\:{numbers}\:{x}, \\ $$$${y}\:{and}\:{z}\:{which}\:{satisfy}\:{the}\:{system} \\ $$$$\begin{cases}{\mathrm{log}\:_{\mathrm{9}} \left({x}\right)+\mathrm{log}\:_{\mathrm{9}} \left({y}\right)+\mathrm{log}\:_{\mathrm{3}} \left({z}\right)=\mathrm{2}}\\{\mathrm{log}\:_{\mathrm{16}} \left({x}\right)+\mathrm{log}\:_{\mathrm{4}} \left({y}\right)+\mathrm{log}\:_{\mathrm{16}} \left({z}\right)=\mathrm{1}}\\{\mathrm{log}\:_{\mathrm{5}} \left({x}\right)+\mathrm{log}\:_{\mathrm{25}} \left({y}\right)+\mathrm{log}\:_{\mathrm{25}} \left({z}\right)=\mathrm{0}}\end{cases}…
Question Number 50085 by F_Nongue last updated on 13/Dec/18 $$\mathrm{5}^{{x}+\mathrm{2}} −\mathrm{95}^{{x}} =\mathrm{2}^{{x}+\mathrm{9}} +\mathrm{1132}^{{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 181154 by mathlove last updated on 22/Nov/22 $$\left({log}_{\mathrm{15}} \mathrm{5}\right)^{\mathrm{2}} +\left({log}_{\mathrm{15}} \mathrm{3}\right)\left({log}_{\mathrm{15}} \mathrm{75}\right)=? \\ $$$$ \\ $$ Answered by SEKRET last updated on 22/Nov/22…
Question Number 50080 by F_Nongue last updated on 13/Dec/18 $$\left.{a}\right)\:{if}\:{f}\left({x}\right)={log}\left({x}+\mathrm{2}\right),\:{solve}\:{the}\:{equation}: \\ $$$$\mathrm{2}^{{f}\left({x}−\mathrm{2}\right)} ×\mathrm{2}^{{f}\left(\mathrm{2}{x}+\mathrm{2}\right)} =\mathrm{4}^{{logf}\left({x}\right)} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Dec/18 $${f}\left({x}−\mathrm{2}\right)={log}\left({x}−\mathrm{2}+\mathrm{2}\right)={logx} \\…
Question Number 115341 by bemath last updated on 25/Sep/20 $${If}\:\mathrm{log}\:\mathrm{tan}\:\mathrm{1}°+\mathrm{log}\:\mathrm{tan}\:\mathrm{2}°+\mathrm{log}\:\mathrm{tan}\:\mathrm{3}°+…+\mathrm{log}\:\mathrm{tan}\:\mathrm{89}°={p} \\ $$$${then}\:{p}^{\mathrm{2}} +\mathrm{3}\:=\: \\ $$ Answered by bobhans last updated on 25/Sep/20 $$\Rightarrow\mathrm{log}\:\left(\mathrm{tan}\:\mathrm{1}°×\mathrm{tan}\:\mathrm{2}°×\mathrm{tan}\:\mathrm{3}°×…×\mathrm{tan}\:\mathrm{89}°\right)={p} \\ $$$${consider}\:\mathrm{tan}\:\mathrm{89}°×\mathrm{tan}\:\mathrm{1}°=\mathrm{1}…
Question Number 115268 by bobhans last updated on 24/Sep/20 $$\sqrt{\mathrm{4}^{{x}} −\mathrm{5}.\mathrm{2}^{{x}+\mathrm{1}} +\mathrm{25}}\:+\sqrt{\mathrm{9}^{{x}} −\mathrm{2}.\mathrm{3}^{{x}+\mathrm{2}} +\mathrm{17}}\:\leqslant\:\mathrm{2}^{{x}} −\mathrm{5} \\ $$ Answered by john santu last updated on 24/Sep/20…