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Question Number 53912 by Mikael_Marshall last updated on 27/Jan/19

{ ((2^x −2^y =1)),((4^x −4^y =(5/3))) :}

$$\begin{cases}{\mathrm{2}^{{x}} −\mathrm{2}^{{y}} =\mathrm{1}}\\{\mathrm{4}^{{x}} −\mathrm{4}^{{y}} =\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$

Answered by byaw last updated on 27/Jan/19

2^x =1+2^y 2^(2x) =(1+2^y )^2 4^x =1+2.2^y +4^y 4^x −4^y =1+2.2^y 1+2.2^y =5/3 3+6.2^y =5 6.2^y =2 2^y =1/3 y=log _2 (1/3) 2^x −2^(log _2 (1/3)) =1 2^x −1/3=1 2^x =4/3 x=log _2 (4/3) x=2−log _2 (1/3)

$$\mathrm{2}^{{x}} =\mathrm{1}+\mathrm{2}^{{y}} \\ $$$$\mathrm{2}^{\mathrm{2}{x}} =\left(\mathrm{1}+\mathrm{2}^{{y}} \right)^{\mathrm{2}} \\ $$$$\mathrm{4}^{{x}} =\mathrm{1}+\mathrm{2}.\mathrm{2}^{{y}} +\mathrm{4}^{{y}} \\ $$$$\mathrm{4}^{{x}} −\mathrm{4}^{{y}} =\mathrm{1}+\mathrm{2}.\mathrm{2}^{{y}} \\ $$$$\mathrm{1}+\mathrm{2}.\mathrm{2}^{{y}} =\mathrm{5}/\mathrm{3} \\ $$$$\mathrm{3}+\mathrm{6}.\mathrm{2}^{{y}} =\mathrm{5} \\ $$$$\mathrm{6}.\mathrm{2}^{{y}} =\mathrm{2} \\ $$$$\mathrm{2}^{{y}} =\mathrm{1}/\mathrm{3} \\ $$$${y}=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{1}/\mathrm{3}\right) \\ $$$$\mathrm{2}^{{x}} −\mathrm{2}^{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{1}/\mathrm{3}\right)} =\mathrm{1} \\ $$$$\mathrm{2}^{{x}} −\mathrm{1}/\mathrm{3}=\mathrm{1} \\ $$$$\mathrm{2}^{{x}} =\mathrm{4}/\mathrm{3} \\ $$$${x}=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{4}/\mathrm{3}\right) \\ $$$${x}=\mathrm{2}−\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{1}/\mathrm{3}\right) \\ $$

Commented by Mikael_Marshall last updated on 27/Jan/19

thanks Sir!

$${thanks}\:{Sir}! \\ $$

Commented by JDamian last updated on 27/Jan/19

Hi, byaw. Your last line is wrong. It should be x=2−log_2 3

$$ \\ $$$$\mathrm{Hi},\:\boldsymbol{\mathrm{byaw}}.\:\mathrm{Your}\:\mathrm{last}\:\mathrm{line}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{It}\:\mathrm{should}\:\mathrm{be} \\ $$$$\:\:\:{x}=\mathrm{2}−\mathrm{log}_{\mathrm{2}} \mathrm{3} \\ $$

Commented by byaw last updated on 15/Feb/19

yes I ve realized

$${yes}\:{I}\:{ve}\:{realized} \\ $$

Answered by peter frank last updated on 27/Jan/19

a−b=1 a^2 −b^2 =(5/3) (a−b)(a+b)=(5/3) a+b=(5/3) 2a=(5/3)+1=(8/3) a=(8/6) −2b=1−(5/3) 2b=(5/3)−1 b=(2/6)

$${a}−{b}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${a}+{b}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{2}{a}=\frac{\mathrm{5}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${a}=\frac{\mathrm{8}}{\mathrm{6}} \\ $$$$−\mathrm{2}{b}=\mathrm{1}−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{2}{b}=\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{1} \\ $$$${b}=\frac{\mathrm{2}}{\mathrm{6}} \\ $$$$ \\ $$

Commented by Mikael_Marshall last updated on 27/Jan/19

thanks Sir

$${thanks}\:{Sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

a−b=1 a^2 −b^2 =(5/3) a+b=(5/3) a−b=1 2a=1+(5/3) a=(4/3) b=(5/3)−(4/3)=(1/3) 2^x =(4/3) x=ln_2 ((4/3)) 2^y =(1/3) y=ln_2 ((1/3))

$${a}−{b}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${a}+{b}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${a}−{b}=\mathrm{1} \\ $$$$\mathrm{2}{a}=\mathrm{1}+\frac{\mathrm{5}}{\mathrm{3}}\:\:\:{a}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${b}=\frac{\mathrm{5}}{\mathrm{3}}−\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{2}^{{x}} =\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${x}={ln}_{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$$$\mathrm{2}^{{y}} =\frac{\mathrm{1}}{\mathrm{3}}\:\:{y}={ln}_{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$

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