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Question Number 168247 by Mastermind last updated on 07/Apr/22

Solve for x ((8^x +27^x )/(12^x +18^x ))=(7/6) Mastermind

$${Solve}\:{for}\:{x} \\ $$$$\frac{\mathrm{8}^{{x}} +\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} +\mathrm{18}^{{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by MJS_new last updated on 07/Apr/22

((8^x +27^x )/(12^x +18^x ))=(7/6) (((2^x +3^x )(4^x −6^x +9^x ))/((2^x +3^x )6^x ))=(7/6) 6×4^x −13×6^x +6×9^x =0 obviously 6×4−13×6+6×9=0 ⇒ x=1 less obviously (6/4)−((13)/6)+(6/9)=0 ⇒ x=−1

$$\frac{\mathrm{8}^{{x}} +\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} +\mathrm{18}^{{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\frac{\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)\left(\mathrm{4}^{{x}} −\mathrm{6}^{{x}} +\mathrm{9}^{{x}} \right)}{\left(\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \right)\mathrm{6}^{{x}} }=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\mathrm{6}×\mathrm{4}^{{x}} −\mathrm{13}×\mathrm{6}^{{x}} +\mathrm{6}×\mathrm{9}^{{x}} =\mathrm{0} \\ $$$$\mathrm{obviously}\:\mathrm{6}×\mathrm{4}−\mathrm{13}×\mathrm{6}+\mathrm{6}×\mathrm{9}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{1} \\ $$$$\mathrm{less}\:\mathrm{obviously} \\ $$$$\frac{\mathrm{6}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{6}}+\frac{\mathrm{6}}{\mathrm{9}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\mathrm{1} \\ $$

Commented by Mastermind last updated on 07/Apr/22

Thanks, please more explanation

$$ \\ $$$${Thanks},\:{please}\:{more}\:{explanation} \\ $$

Commented by MJS_new last updated on 07/Apr/22

sorry I can′t explain why this is obvious to me

$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{explain}\:\mathrm{why}\:\mathrm{this}\:\mathrm{is}\:\mathrm{obvious}\:\mathrm{to}\:\mathrm{me} \\ $$

Commented by Mastermind last updated on 07/Apr/22

Okay, Anyways, Thanks so much

$${Okay}, \\ $$$${Anyways},\:{Thanks}\:{so}\:{much} \\ $$

Answered by cortano1 last updated on 07/Apr/22

((8^x +27^x )/(12^x +18^x )) = (7/6) ⇒(((2^x )^3 +(3^x )^3 )/((2^x )^2 (3^x )+(3^x )^2 (2^x ))) = (7/6) ⇒ ((((2^x /3^x ))^3 +1)/(((2^x /3^x ))^2 +((2^x /3^x )))) = (7/6) let ((2/3))^x = d ; d>0 ⇒ ((d^3 +1)/(d^2 +d)) = (7/6) ⇒(((d+1)(d^2 −d+1))/(d(d+1))) = (7/6) ⇒6d^2 −6d+6= 7d ⇒6d^2 −13d+6=0 ⇒(2d−3)(3d−2)=0 ⇒ { ((d=(3/2)⇒((2/3))^x =((2/3))^(−1) ; x=−1)),((d=(2/3)⇒((2/3))^x =((2/3))^1 ; x=1)) :} check : x=1⇒((8+27)/(12+18))=((35)/(30))=(7/6) x=−1⇒(((1/(18))+(1/(27)))/((1/(12))+(1/(18)))) = (7/6)

$$\:\:\frac{\mathrm{8}^{{x}} +\mathrm{27}^{{x}} }{\mathrm{12}^{{x}} +\mathrm{18}^{{x}} }\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\:\Rightarrow\frac{\left(\mathrm{2}^{{x}} \right)^{\mathrm{3}} +\left(\mathrm{3}^{{x}} \right)^{\mathrm{3}} }{\left(\mathrm{2}^{{x}} \right)^{\mathrm{2}} \left(\mathrm{3}^{{x}} \right)+\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} \left(\mathrm{2}^{{x}} \right)}\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\Rightarrow\:\frac{\left(\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}} }\right)^{\mathrm{3}} +\mathrm{1}}{\left(\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}} }\right)}\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\:{let}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \:=\:{d}\:;\:{d}>\mathrm{0} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{3}} +\mathrm{1}}{{d}^{\mathrm{2}} +{d}}\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\left({d}+\mathrm{1}\right)\left({d}^{\mathrm{2}} −{d}+\mathrm{1}\right)}{{d}\left({d}+\mathrm{1}\right)}\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{6}{d}^{\mathrm{2}} −\mathrm{6}{d}+\mathrm{6}=\:\mathrm{7}{d} \\ $$$$\Rightarrow\mathrm{6}{d}^{\mathrm{2}} −\mathrm{13}{d}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}{d}−\mathrm{3}\right)\left(\mathrm{3}{d}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{d}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{−\mathrm{1}} ;\:{x}=−\mathrm{1}}\\{{d}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{1}} ;\:{x}=\mathrm{1}}\end{cases} \\ $$$$\:{check}\::\:{x}=\mathrm{1}\Rightarrow\frac{\mathrm{8}+\mathrm{27}}{\mathrm{12}+\mathrm{18}}=\frac{\mathrm{35}}{\mathrm{30}}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$$$\:{x}=−\mathrm{1}\Rightarrow\frac{\frac{\mathrm{1}}{\mathrm{18}}+\frac{\mathrm{1}}{\mathrm{27}}}{\frac{\mathrm{1}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{18}}}\:=\:\frac{\mathrm{7}}{\mathrm{6}} \\ $$

Commented by peter frank last updated on 07/Apr/22

good

$$\mathrm{good} \\ $$

Commented by Mastermind last updated on 07/Apr/22

you did a great Job

$$ \\ $$$${you}\:{did}\:{a}\:{great}\:{Job} \\ $$

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