Question Number 145437 by Saiki last updated on 04/Jul/21 $${original}\:{length}\:{of}\:{the}\:{iron}\:{rod}=\mathrm{175}.\mathrm{65} \\ $$$$\%\:{increase}=\mathrm{6}\frac{\mathrm{1}}{\mathrm{3}}\%×\mathrm{175}.\mathrm{65} \\ $$$$=\frac{\mathrm{19}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{100}}×\mathrm{175}.\mathrm{65} \\ $$$$=\frac{\mathrm{19}×\mathrm{175}.\mathrm{65}}{\mathrm{3}×\mathrm{100}}=\frac{\mathrm{3337}.\mathrm{35}}{\mathrm{300}}=\mathrm{11}.\mathrm{1245} \\ $$$${new}\:{length}={original}\:{length}+{increased}\:{length} \\ $$$$=\mathrm{175}.\mathrm{65}+\mathrm{11}.\mathrm{1245} \\ $$$$=\mathrm{186}.\mathrm{7745}{cm} \\ $$$${solution}\:{by}\:{CASIO}….. \\…
Question Number 145108 by imjagoll last updated on 02/Jul/21 Answered by liberty last updated on 02/Jul/21 $$\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\sqrt[{\mathrm{1}/\mathrm{5}}]{\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{32}^{\mathrm{2}{x}−\mathrm{2}} }}}=\:\mathrm{64} \\ $$$$\Rightarrow\:\sqrt[{\mathrm{1}/\mathrm{4}}]{\:\left(\frac{\mathrm{2}^{{x}+\mathrm{1}} }{\mathrm{2}^{\mathrm{10}{x}−\mathrm{10}} }\right)^{\mathrm{5}} }=\:\mathrm{2}^{\mathrm{6}} \\…
Question Number 145071 by imjagoll last updated on 02/Jul/21 $$\mathrm{log}\:_{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \left(\mathrm{x}+\mathrm{2}\right)\:=\:\mathrm{1}+\:\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{4}−\mathrm{x}\right) \\ $$$$\:\mathrm{x}=? \\ $$ Answered by MJS_new last updated on 02/Jul/21 $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\…
Question Number 145067 by imjagoll last updated on 02/Jul/21 $$\:\mathrm{log}\:_{\mathrm{3}} \left({x}+\mathrm{1}\right)\:=\mathrm{log}\:_{\mathrm{4}} \left({x}+\mathrm{8}\right) \\ $$$$\:{x}=? \\ $$ Answered by bobhans last updated on 02/Jul/21 $$\:\mathrm{x}=\mathrm{8}\:\Rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{8}+\mathrm{1}\right)=\mathrm{log}\:_{\mathrm{4}}…
Question Number 79263 by jagoll last updated on 23/Jan/20 $$\mathrm{4}^{\mathrm{2x}−\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{4}}\:^{\mathrm{2}} \mathrm{log}^{\mathrm{2}} \left(\mathrm{2x}\right)>\:^{\mathrm{2}} \mathrm{log}\left(\mathrm{x}\right) \\ $$$$\left\{^{\mathrm{2}} \mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{2}^{\mathrm{2x}} \right\} \\ $$ Commented by john santu last…
Question Number 144504 by alcohol last updated on 26/Jun/21 $${for}\:{what}\:{values}\:{of}\:{b}\:{is} \\ $$$${ln}\left({a}−{b}\right){ln}\left({a}+{b}\right)\leqslant{ln}^{\mathrm{2}} {a} \\ $$$${Note}:\:\mathrm{0}\leqslant{b}<{a} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 13152 by tawa tawa last updated on 15/May/17 $$\mathrm{Solve}:\:\:\mathrm{5}^{\mathrm{log}\left(\mathrm{x}\right)} \:+\:\mathrm{logx}^{\mathrm{5}} \:=\:\mathrm{25} \\ $$ Answered by mrW1 last updated on 16/May/17 $$\mathrm{log}\:{x}={t} \\ $$$$\mathrm{5}^{{t}}…
Question Number 13151 by Tinkutara last updated on 15/May/17 Answered by mrW1 last updated on 16/May/17 $${x}>\mathrm{9} \\ $$$$\mathrm{log}\:\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{2} \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)\left({x}−\mathrm{9}\right)>\mathrm{100} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{19}{x}−\mathrm{91}>\mathrm{0} \\…
Question Number 65934 by gunawan last updated on 06/Aug/19 $$\:\mathrm{log}_{\mathrm{7}} \:\mathrm{2}={a} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\mathrm{3}={b} \\ $$$$\:\mathrm{log}_{\mathrm{6}} \:\mathrm{98}=… \\ $$ Answered by meme last updated on…
Question Number 65930 by gunawan last updated on 06/Aug/19 $$\mathrm{If}\:{t}=\frac{{x}^{\mathrm{2}} −\mathrm{3}}{\mathrm{3}{x}+\mathrm{7}}\:\mathrm{then} \\ $$$$\mathrm{log}\left(\mathrm{1}−\mid{t}\mid\right)\:\mathrm{can}\:\mathrm{to}\:\mathrm{find}\:\mathrm{for} \\ $$$$\mathrm{a}.\:\mathrm{2}<{x}<\mathrm{6} \\ $$$${b}.−\:\mathrm{2}<{x}<\mathrm{5} \\ $$$${c}.−\:\mathrm{2}\leqslant{x}\leqslant\mathrm{6} \\ $$$$\mathrm{d}.\:{x}\leqslant−\mathrm{2}\:\mathrm{or}\:{x}>\mathrm{6} \\ $$$${e}.\:{x}<−\mathrm{1}\:{or}\:{x}>\mathrm{3} \\ $$…