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Question Number 19063 by chux last updated on 03/Aug/17

find the possible values of x if ((8^x +27^x )/(12^x +18^x ))=(7/6)

$$\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{if} \\ $$$$\frac{\mathrm{8}^{\mathrm{x}} +\mathrm{27}^{\mathrm{x}} }{\mathrm{12}^{\mathrm{x}} +\mathrm{18}^{\mathrm{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$

Answered by 433 last updated on 03/Aug/17

((2^(3x) +3^(3x) )/((3×4)^x +(9×2)^x ))=(7/6) ((2^(3x) +3^(3x) )/(6^x (2^x +3^x )))=(7/6) ((2^(3x) +3^(3x) )/((2×3)^x (2^x +3^x )))=(7/6) 2^x =u & 3^x =v ((u^3 +v^3 )/(uv(u+v)))=(7/6) 6u^3 +6v^3 =7u^2 v+7uv^2 6(u+v)^3 =25uv(u+v) (u+v)(6(u+v)^2 −25uv)=0 u+v=0 ⇔ 2^x +3^x =0 ∗ 6(u+v)^2 −25uv=0 6u^2 +12uv+6v^2 −25uv=0 6u^2 −13uv+6v^2 =0 6u^2 −4uv−9uv+6v^2 =0 2u(3u−2v)−3v(3u−2v)=0 (2u−3v)(3u−2v)=0 2u=3v ⇔ 2×2^x =3×3^x ⇔ 2^(x+1) =3^(x+1) ⇔ ((2/3))^(x+1) =((2/3))^0 ⇔ x=−1 3u=2v ⇔ 3×2^x =2×3^x ⇔ 2^(x−1) =3^(x−1) ⇔ ((2/3))^(x−1) =((2/3))^0 ⇔ x=1

$$ \\ $$$$ \\ $$$$\frac{\mathrm{2}^{\mathrm{3x}} +\mathrm{3}^{\mathrm{3x}} }{\left(\mathrm{3}×\mathrm{4}\right)^{\mathrm{x}} +\left(\mathrm{9}×\mathrm{2}\right)^{\mathrm{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\frac{\mathrm{2}^{\mathrm{3x}} +\mathrm{3}^{\mathrm{3x}} }{\mathrm{6}^{\mathrm{x}} \left(\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \right)}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\frac{\mathrm{2}^{\mathrm{3x}} +\mathrm{3}^{\mathrm{3x}} }{\left(\mathrm{2}×\mathrm{3}\right)^{\mathrm{x}} \left(\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \right)}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\mathrm{2}^{\mathrm{x}} =\mathrm{u}\:\&\:\mathrm{3}^{\mathrm{x}} =\mathrm{v} \\ $$$$\frac{\mathrm{u}^{\mathrm{3}} +\mathrm{v}^{\mathrm{3}} }{\mathrm{uv}\left(\mathrm{u}+\mathrm{v}\right)}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\mathrm{6u}^{\mathrm{3}} +\mathrm{6v}^{\mathrm{3}} =\mathrm{7u}^{\mathrm{2}} \mathrm{v}+\mathrm{7uv}^{\mathrm{2}} \\ $$$$\mathrm{6}\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{3}} =\mathrm{25uv}\left(\mathrm{u}+\mathrm{v}\right) \\ $$$$\left(\mathrm{u}+\mathrm{v}\right)\left(\mathrm{6}\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}} −\mathrm{25uv}\right)=\mathrm{0} \\ $$$$\mathrm{u}+\mathrm{v}=\mathrm{0}\:\Leftrightarrow\:\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} =\mathrm{0}\:\:\ast \\ $$$$\mathrm{6}\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}} −\mathrm{25uv}=\mathrm{0} \\ $$$$\mathrm{6u}^{\mathrm{2}} +\mathrm{12uv}+\mathrm{6v}^{\mathrm{2}} −\mathrm{25uv}=\mathrm{0} \\ $$$$\mathrm{6u}^{\mathrm{2}} −\mathrm{13uv}+\mathrm{6v}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{6u}^{\mathrm{2}} −\mathrm{4uv}−\mathrm{9uv}+\mathrm{6v}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2u}\left(\mathrm{3u}−\mathrm{2v}\right)−\mathrm{3v}\left(\mathrm{3u}−\mathrm{2v}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2u}−\mathrm{3v}\right)\left(\mathrm{3u}−\mathrm{2v}\right)=\mathrm{0} \\ $$$$\mathrm{2u}=\mathrm{3v}\:\Leftrightarrow\:\mathrm{2}×\mathrm{2}^{\mathrm{x}} =\mathrm{3}×\mathrm{3}^{\mathrm{x}} \:\Leftrightarrow\:\mathrm{2}^{\mathrm{x}+\mathrm{1}} =\mathrm{3}^{\mathrm{x}+\mathrm{1}} \:\Leftrightarrow\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}+\mathrm{1}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \:\Leftrightarrow\:\mathrm{x}=−\mathrm{1} \\ $$$$\mathrm{3u}=\mathrm{2v}\:\Leftrightarrow\:\mathrm{3}×\mathrm{2}^{\mathrm{x}} =\mathrm{2}×\mathrm{3}^{\mathrm{x}} \:\Leftrightarrow\:\mathrm{2}^{\mathrm{x}−\mathrm{1}} =\mathrm{3}^{\mathrm{x}−\mathrm{1}} \:\Leftrightarrow\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{x}−\mathrm{1}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{0}} \:\Leftrightarrow\:\mathrm{x}=\mathrm{1} \\ $$

Commented by NEC last updated on 04/Aug/17

hmmmm

$${hmmmm} \\ $$

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

sir pls i dnt understan how u got that 6^x

$${sir}\:{pls}\:{i}\:{dnt}\:{understan}\:{how}\:{u}\:{got}\:{that}\:\mathrm{6}^{{x}} \\ $$

Commented by 433 last updated on 04/Aug/17

(3×4)^x +(9×2)^x =3^x 2^x 2^x +3^x 3^x 2^x =3^x 2^x (2^x +3^x )=6^x (2^x +3^x )

$$ \\ $$$$\left(\mathrm{3}×\mathrm{4}\right)^{\mathrm{x}} +\left(\mathrm{9}×\mathrm{2}\right)^{\mathrm{x}} =\mathrm{3}^{\mathrm{x}} \mathrm{2}^{\mathrm{x}} \mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \mathrm{3}^{\mathrm{x}} \mathrm{2}^{\mathrm{x}} =\mathrm{3}^{\mathrm{x}} \mathrm{2}^{\mathrm{x}} \left(\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \right)=\mathrm{6}^{\mathrm{x}} \left(\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} \right) \\ $$

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

waaw brilliant sir thankz i comply now

$${waaw}\:{brilliant}\:{sir}\:{thankz}\:{i}\:{comply}\:{now} \\ $$

Answered by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17

2^x =a,3^x =b,a+b≠0⇒((a^3 +b^3 )/(a^2 b+ab^2 ))=(7/6)⇒ 6(a^2 −ab+b^2 )=7ab⇒6a^2 −13ab+6b^2 =0 a=((13b±(√(169b^2 −144b^2 )))/(12))=((13b±5b)/(12))=((3/2),(2/3))b ⇒ { (((a/b)=(2/3)⇒((2/3))^x =(2/3)⇒x=1)),(((a/b)=(3/2)⇒((2/3))^x =(3/2)⇒x=−1 .)) :}

$$\mathrm{2}^{{x}} ={a},\mathrm{3}^{{x}} ={b},{a}+{b}\neq\mathrm{0}\Rightarrow\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{6}}\Rightarrow \\ $$$$\mathrm{6}\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)=\mathrm{7}{ab}\Rightarrow\mathrm{6}{a}^{\mathrm{2}} −\mathrm{13}{ab}+\mathrm{6}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}=\frac{\mathrm{13}{b}\pm\sqrt{\mathrm{169}{b}^{\mathrm{2}} −\mathrm{144}{b}^{\mathrm{2}} }}{\mathrm{12}}=\frac{\mathrm{13}{b}\pm\mathrm{5}{b}}{\mathrm{12}}=\left(\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{2}}{\mathrm{3}}\right){b} \\ $$$$\Rightarrow\begin{cases}{\frac{{a}}{{b}}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow{x}=\mathrm{1}}\\{\frac{{a}}{{b}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow{x}=−\mathrm{1}\:.}\end{cases} \\ $$

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

hmm sir am baffle about how u got that 6(a^2 −ab+b^2 )

$${hmm}\:{sir}\:{am}\:{baffle}\:{about}\:{how}\:{u}\:{got}\:{that}\:\mathrm{6}\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$$ \\ $$

Commented by behi.8.3.4.1.7@gmail.com last updated on 04/Aug/17

((a^3 +b^3 )/(a^2 b+ab^2 ))=(((a+b)(a^2 −ab+b^2 ))/(ab(a+b)))=((a^2 −ab+b^2 )/(ab))

$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} }=\frac{\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right)}{{ab}\left({a}+{b}\right)}=\frac{{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} }{{ab}} \\ $$

Commented by chernoaguero@gmail.com last updated on 04/Aug/17

Thankx sir

$${Thankx}\:{sir}\: \\ $$

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