Question Number 136520 by mey3nipaba last updated on 22/Mar/21
Commented by mey3nipaba last updated on 22/Mar/21
$$\mathrm{help}\:\mathrm{please}. \\ $$
Answered by mindispower last updated on 22/Mar/21
$$\Leftrightarrow\begin{cases}{\mathrm{2}^{{x}} =\mathrm{3}^{{y}} \Rightarrow{x}=\frac{{yln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)}}\\{\left({x}−\mathrm{1}\right)=\frac{{y}−\mathrm{1}}{{ln}\left(\mathrm{3}\right)}{ln}\left(\mathrm{2}\right)}\end{cases} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 23/Mar/21
$$\mathrm{2}^{{x}+{y}} =\mathrm{6}^{{y}} \:\wedge\:\mathrm{3}^{{x}} =\mathrm{3}\left(\mathrm{2}^{{y}−\mathrm{1}} \right) \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} ….\left({i}\right)\:\wedge\:\mathrm{3}^{{x}−\mathrm{1}} =\mathrm{2}^{{y}−\mathrm{1}} ….\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\mathrm{2}^{{x}+{y}−\mathrm{1}} =\mathrm{3}^{{x}+{y}−\mathrm{1}} \\ $$$$\Rightarrow{x}+{y}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}+{y}=\mathrm{1}………………\left({iii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{2}^{{y}−\mathrm{1}} }=\frac{\mathrm{3}^{{y}} }{\mathrm{3}^{{x}−\mathrm{1}} } \\ $$$$\mathrm{2}^{{x}−{y}+\mathrm{1}} =\mathrm{3}^{{y}−{x}+\mathrm{1}} \\ $$$${x}−{y}+\mathrm{1}=\mathrm{0}\:\wedge\:{y}−{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}−{y}+\mathrm{1}={y}−{x}+\mathrm{1} \\ $$$${x}−{y}={y}−{x} \\ $$$${x}={y}……………………..\left({iv}\right) \\ $$$$\left({iii}\right)\:\&\:\left({iv}\right): \\ $$$${x}={y}=\mathrm{1}/\mathrm{2} \\ $$$${Doesn}'{t}\:{satisfy}\:{original}\:{equation}. \\ $$$$\therefore\:{No}\:{solution}. \\ $$
Commented by bemath last updated on 23/Mar/21
$${if}\:{x}={y}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}^{{x}+{y}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{2}^{\mathrm{1}} \\ $$$${RHS}\Rightarrow\mathrm{6}^{{y}} \:=\:\mathrm{6}^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{6}} \\ $$$${why}\:\mathrm{2}\neq\sqrt{\mathrm{6}}\:{sir}? \\ $$
Commented by Rasheed.Sindhi last updated on 23/Mar/21
$${Yes}\:{madam},\:{perhaps}\:{there}'{s}\:{no}\:{solution}. \\ $$
Commented by mr W last updated on 23/Mar/21
$$\mathrm{3}^{{a}} =\mathrm{2}^{{b}} \:\nRightarrow\:{a}={b}=\mathrm{0} \\ $$$$ \\ $$$${i}\:{think}\:{it}'{s}\:{wrong}\:{here}: \\ $$$$\mathrm{2}^{{x}−{y}+\mathrm{1}} =\mathrm{3}^{{y}−{x}+\mathrm{1}} \\ $$$$\nRightarrow{x}−{y}+\mathrm{1}=\mathrm{0}\:\wedge\:{y}−{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}−{y}+\mathrm{1}=\left({y}−{x}+\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} \:\mathrm{3} \\ $$$$\Rightarrow{x}−{y}=\frac{\mathrm{log}_{\mathrm{2}} \:\mathrm{3}}{\mathrm{1}+\mathrm{log}_{\mathrm{2}} \:\mathrm{3}} \\ $$
Commented by Rasheed.Sindhi last updated on 23/Mar/21
$$\mathcal{T}{h}\alpha{n}\Bbbk{s}\:{for}\:{guidline}\:\boldsymbol{{sir}}! \\ $$
Answered by mr W last updated on 23/Mar/21
$$\mathrm{2}^{{x}} \mathrm{2}^{{y}} =\mathrm{2}^{{y}} \mathrm{3}^{{y}} \\ $$$$\mathrm{2}^{{x}} =\mathrm{3}^{{y}} \\ $$$$\Rightarrow{x}={y}\mathrm{log}_{\mathrm{2}} \:\mathrm{3}={ky} \\ $$$${with}\:{k}=\mathrm{log}_{\mathrm{2}} \:\mathrm{3} \\ $$$$\mathrm{3}^{{x}−\mathrm{1}} =\mathrm{2}^{{y}−\mathrm{1}} \\ $$$${y}−\mathrm{1}=\left({x}−\mathrm{1}\right)\mathrm{log}_{\mathrm{2}} \:\mathrm{3}=\left({x}−\mathrm{1}\right){k} \\ $$$${y}−\mathrm{1}=\left({ky}−\mathrm{1}\right){k} \\ $$$$\Rightarrow{y}=\frac{\mathrm{1}−{k}}{\mathrm{1}−{k}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{1}+{k}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{log}_{\mathrm{2}} \:\mathrm{3}}\approx\mathrm{0}.\mathrm{386853} \\ $$$$\Rightarrow{x}=\frac{{k}}{\mathrm{1}+{k}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{k}}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{log}_{\mathrm{3}} \:\mathrm{2}}\approx\mathrm{0}.\mathrm{613147} \\ $$