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

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

Answered by 433 last updated on 03/Aug/17

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

Commented by NEC last updated on 04/Aug/17

$$hmmmm$$

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

$$sirplsidntunderstanhowugotthat{6}^x$$

Commented by 433 last updated on 04/Aug/17

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

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

$$waawbrilliantsirthankzicomplynow$$

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

$$2^x=a,3^x=b,a+b\ne 0\Rightarrow \frac{a^3+b^3}{a^{2}b+{ab}^2}=\frac{7}{6}\Rightarrow$$$$6(a^2-ab+b^2)=7ab\Rightarrow 6a^2-13ab+6b^2=0$$$$a=\frac{13b\pm \sqrt{169b^2-144b^2}}{12}=\frac{13b\pm 5b}{12}=(\frac{3}{2},\frac{2}{3})b$$$$\Rightarrow \{\begin{array}{l}\frac{a}{b}=\frac{2}{3}\Rightarrow {\left(\frac{2}{3}\right)}^x=\frac{2}{3}\Rightarrow x=1\\ \frac{a}{b}=\frac{3}{2}\Rightarrow {\left(\frac{2}{3}\right)}^x=\frac{3}{2}\Rightarrow x=-1.\end{array}$$

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

$$hmmsirambaffleabouthowugotthat6(a^2-ab+b^2)$$$$$$

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

$$\frac{a^3+b^3}{a^{2}b+{ab}^2}=\frac{(a+b)(a^2-ab+b^2)}{ab(a+b)}=\frac{a^2-ab+b^2}{ab}$$

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

$$Thankxsir$$

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