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2-x-3-y-6-3-x-4-y-12-




Question Number 87439 by mary_ last updated on 04/Apr/20
 { ((2^x .3^y =6)),((3^x .4^y =12)) :}
$$\begin{cases}{\mathrm{2}^{\mathrm{x}} .\mathrm{3}^{\mathrm{y}} =\mathrm{6}}\\{\mathrm{3}^{\mathrm{x}} .\mathrm{4}^{\mathrm{y}} =\mathrm{12}}\end{cases} \\ $$
Commented by john santu last updated on 04/Apr/20
((2^x .3^y )/(4^y .3^x )) = (1/2) ⇒ 3^(y−x)  = (2^(2y) /2^(x+1) )  3^(y−x)  = 2^(2y−x−1)   y−x = (2y−x−1) log_3 (2)...(i)  2^(x−1) . 3^y  = 3 ⇒ 2^(x−1)  = 3^(1−y)   (x−1) log_3  (2)= 1−y ... (2)
$$\frac{\mathrm{2}^{\mathrm{x}} .\mathrm{3}^{\mathrm{y}} }{\mathrm{4}^{\mathrm{y}} .\mathrm{3}^{\mathrm{x}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{3}^{\mathrm{y}−\mathrm{x}} \:=\:\frac{\mathrm{2}^{\mathrm{2y}} }{\mathrm{2}^{\mathrm{x}+\mathrm{1}} } \\ $$$$\mathrm{3}^{\mathrm{y}−\mathrm{x}} \:=\:\mathrm{2}^{\mathrm{2y}−\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{y}−\mathrm{x}\:=\:\left(\mathrm{2y}−\mathrm{x}−\mathrm{1}\right)\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}\right)…\left(\mathrm{i}\right) \\ $$$$\mathrm{2}^{\mathrm{x}−\mathrm{1}} .\:\mathrm{3}^{\mathrm{y}} \:=\:\mathrm{3}\:\Rightarrow\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{1}−\mathrm{y}} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)\:\mathrm{log}_{\mathrm{3}} \:\left(\mathrm{2}\right)=\:\mathrm{1}−\mathrm{y}\:…\:\left(\mathrm{2}\right) \\ $$
Commented by mr W last updated on 04/Apr/20
2^(x−1) 3^(y−1) =1 ⇒y−1=(((x−1)ln 2)/(ln 3))  3^(x−1) 4^(y−1) =1 ⇒y−1=(((x−1)ln 3)/(ln 4))  (x−1)[((ln 2)/(ln 3))−((ln 3)/(ln 4))]=0  ⇒x−1=0 ⇒x=1  ⇒y−1=0 ⇒y=1
$$\mathrm{2}^{{x}−\mathrm{1}} \mathrm{3}^{{y}−\mathrm{1}} =\mathrm{1}\:\Rightarrow{y}−\mathrm{1}=\frac{\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}} \\ $$$$\mathrm{3}^{{x}−\mathrm{1}} \mathrm{4}^{{y}−\mathrm{1}} =\mathrm{1}\:\Rightarrow{y}−\mathrm{1}=\frac{\left({x}−\mathrm{1}\right)\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{4}} \\ $$$$\left({x}−\mathrm{1}\right)\left[\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{4}}\right]=\mathrm{0} \\ $$$$\Rightarrow{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow{y}−\mathrm{1}=\mathrm{0}\:\Rightarrow{y}=\mathrm{1} \\ $$

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