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Question Number 83759 by M±th+et£s last updated on 05/Mar/20
3^x 8^(x/(x+2)) =6
$$\mathrm{3}^{{x}} \:\mathrm{8}^{\frac{{x}}{{x}+\mathrm{2}}} =\mathrm{6} \\ $$
Commented by M±th+et£s last updated on 05/Mar/20
solve the equation
$${solve}\:{the}\:{equation} \\ $$
Commented by niroj last updated on 06/Mar/20
3^x .8^(x/(x+2)) =6 3^x .2^(3((x/(x+2)))) =6 3^x .2^((3x)/(x+2)) =3.2 3^(x−1) .2^(((3x)/(x+2))−1) =1 3^(x−1) .2^((3x−x−2)/(x+2)) =(3.2)^0 3^(x−1) .2^((2x−2)/(x+2)) =(3.2)^0 3^(x−1) .2^((2(x−1))/(x−1+3)) =1 put x−1=a 3^a .2^((2a)/(a+3)) =3^0 .2^0 3^a =3^0 , 2^((2a)/(a+3)) =2^0 a=0 , ((2a)/(a+3))=0 x−1=0, 2a=0⇒ a=0⇒ x−1=0 x=1. ∴ the value of x is 1.
$$\:\:\mathrm{3}^{{x}} .\mathrm{8}^{\frac{{x}}{{x}+\mathrm{2}}} =\mathrm{6} \\ $$$$\:\:\mathrm{3}^{{x}} .\mathrm{2}^{\mathrm{3}\left(\frac{{x}}{{x}+\mathrm{2}}\right)} =\mathrm{6}\:\: \\ $$$$\:\mathrm{3}^{{x}} .\mathrm{2}^{\frac{\mathrm{3}{x}}{{x}+\mathrm{2}}} =\mathrm{3}.\mathrm{2} \\ $$$$\:\:\:\mathrm{3}^{{x}−\mathrm{1}} .\mathrm{2}^{\frac{\mathrm{3}{x}}{{x}+\mathrm{2}}−\mathrm{1}} =\mathrm{1} \\ $$$$\:\:\mathrm{3}^{{x}−\mathrm{1}} .\mathrm{2}^{\frac{\mathrm{3}{x}−{x}−\mathrm{2}}{{x}+\mathrm{2}}} =\left(\mathrm{3}.\mathrm{2}\right)^{\mathrm{0}} \\ $$$$\:\:\mathrm{3}^{{x}−\mathrm{1}} .\mathrm{2}^{\frac{\mathrm{2}{x}−\mathrm{2}}{{x}+\mathrm{2}}} =\left(\mathrm{3}.\mathrm{2}\right)^{\mathrm{0}} \\ $$$$\:\:\mathrm{3}^{{x}−\mathrm{1}} .\mathrm{2}^{\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{{x}−\mathrm{1}+\mathrm{3}}} =\mathrm{1} \\ $$$$\:\:{put}\:{x}−\mathrm{1}={a} \\ $$$$\:\:\:\mathrm{3}^{{a}} .\mathrm{2}^{\frac{\mathrm{2}{a}}{{a}+\mathrm{3}}} =\mathrm{3}^{\mathrm{0}} .\mathrm{2}^{\mathrm{0}} \\ $$$$\:\:\:\:\mathrm{3}^{{a}} =\mathrm{3}^{\mathrm{0}} ,\:\:\:\mathrm{2}^{\frac{\mathrm{2}{a}}{{a}+\mathrm{3}}} =\mathrm{2}^{\mathrm{0}} \\ $$$$\:\:\:{a}=\mathrm{0}\:,\:\:\:\:\:\:\:\frac{\mathrm{2}{a}}{{a}+\mathrm{3}}=\mathrm{0} \\ $$$$\:\:{x}−\mathrm{1}=\mathrm{0},\:\:\mathrm{2}{a}=\mathrm{0}\Rightarrow\:{a}=\mathrm{0}\Rightarrow\:{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}=\mathrm{1}.\:\: \\ $$$$\:\therefore\:{the}\:{value}\:{of}\:{x}\:{is}\:\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 06/Mar/20
if a^c b^d =1 there are two solutions: solution 1: c=d=0 solution 2: a^c =b^(−d) ⇒c ln a=−d ln b ⇒(c/d)=−((ln b)/(ln a))
$${if}\:{a}^{{c}} {b}^{{d}} =\mathrm{1}\:{there}\:{are}\:\boldsymbol{{two}}\:{solutions}: \\ $$$${solution}\:\mathrm{1}:\:{c}={d}=\mathrm{0} \\ $$$${solution}\:\mathrm{2}:\:{a}^{{c}} ={b}^{−{d}} \:\Rightarrow{c}\:\mathrm{ln}\:{a}=−{d}\:\mathrm{ln}\:{b}\:\Rightarrow\frac{{c}}{{d}}=−\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:{a}} \\ $$
Answered by behi83417@gmail.com last updated on 05/Mar/20
3^(x−1) =2^(1−((3x)/(x+2))) ⇒ { ((x−1=0⇒x=1)),((1−((3x)/(x+2))=0⇒^(x≠−2) 3x=x+2⇒x=1)) :}
$$\mathrm{3}^{\mathrm{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{1}−\frac{\mathrm{3x}}{\mathrm{x}+\mathrm{2}}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{1}}\\{\mathrm{1}−\frac{\mathrm{3x}}{\mathrm{x}+\mathrm{2}}=\mathrm{0}\overset{\mathrm{x}\neq−\mathrm{2}} {\Rightarrow}\:\:\mathrm{3x}=\mathrm{x}+\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{1}}\end{cases} \\ $$
Answered by mr W last updated on 05/Mar/20
3^(x−1) 2^((2(x−1))/(x+2)) =1 [3×2^(2/(x+2)) ]^(x−1) =1 x−1=0 ⇒x=1 or 3×2^(2/(x+2)) =1 2^(2/(x+1)) =(1/3) (2/(x+2))=−((ln 3)/(ln 2)) ⇒x=−(2+((2ln 2)/(ln 3)))≈−3.2619
$$\mathrm{3}^{{x}−\mathrm{1}} \mathrm{2}^{\frac{\mathrm{2}\left({x}−\mathrm{1}\right)}{{x}+\mathrm{2}}} =\mathrm{1} \\ $$$$\left[\mathrm{3}×\mathrm{2}^{\frac{\mathrm{2}}{{x}+\mathrm{2}}} \right]^{{x}−\mathrm{1}} =\mathrm{1} \\ $$$${x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{1} \\ $$$${or} \\ $$$$\mathrm{3}×\mathrm{2}^{\frac{\mathrm{2}}{{x}+\mathrm{2}}} =\mathrm{1} \\ $$$$\mathrm{2}^{\frac{\mathrm{2}}{{x}+\mathrm{1}}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{\mathrm{2}}{{x}+\mathrm{2}}=−\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\Rightarrow{x}=−\left(\mathrm{2}+\frac{\mathrm{2ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{3}}\right)\approx−\mathrm{3}.\mathrm{2619} \\ $$

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