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Question Number 147223 by mathlove last updated on 19/Jul/21

Commented by bramlexs22 last updated on 19/Jul/21

$$2^{3\mathrm{x}}+2^{\mathrm{x}}.3^{2\mathrm{x}}-2.3^{3\mathrm{x}}=0$$$$\{\begin{array}{l}{2}^{\mathrm{x}}=\mathrm{u}\\ 3^{\mathrm{x}}=\mathrm{v}\end{array}\Rightarrow {\mathrm{u}}^3+\mathrm{u}.{\mathrm{v}}^2-{2\mathrm{v}}^3=0$$$$\Rightarrow 1+{\left(\frac{\mathrm{v}}{\mathrm{u}}\right)}^2-2{\left(\frac{\mathrm{v}}{\mathrm{u}}\right)}^3=0$$$$\Rightarrow {2\mathrm{t}}^3-{\mathrm{t}}^2-1=0$$$$\Rightarrow (\mathrm{t}-1)({2\mathrm{t}}^2+\mathrm{t}+1)=0$$$$\mathrm{for}\mathrm{t}=1\Rightarrow \mathrm{u}=\mathrm{v}$$$$\Rightarrow 2^{\mathrm{x}}=3^{\mathrm{x}};{\left(\frac{2}{3}\right)}^{\mathrm{x}}=1;\mathrm{x}=0$$$$\mathrm{for}{2\mathrm{t}}^2+\mathrm{t}+1=0;\mathrm{not}\mathrm{real}\mathrm{for}\mathrm{x}$$$$$$

Answered by puissant last updated on 19/Jul/21

$$\Rightarrow 8^{\mathrm{x}}+{18}^{\mathrm{x}}=2\times {27}^{\mathrm{x}}$$$$\Rightarrow {\left(\frac{8}{18}\right)}^{\mathrm{x}}+1=2{\left(\frac{27}{18}\right)}^{\mathrm{x}}$$$$\Rightarrow {\left(\frac{2}{3}\right)}^{2\mathrm{x}}+1=2{\left(\frac{3}{2}\right)}^{\mathrm{x}}$$$$\mathrm{let}\mathrm{A}={\left(\frac{2}{3}\right)}^{\mathrm{x}}\Rightarrow {\mathrm{A}}^{-1}={\left(\frac{3}{2}\right)}^{\mathrm{x}}$$$$\Rightarrow {\mathrm{A}}^2+1={2\mathrm{A}}^{-1}$$$$\Rightarrow {\mathrm{A}}^3+\mathrm{A}-2=0$$$$\Rightarrow (\mathrm{A}-1)({\mathrm{A}}^2+\mathrm{A}+2)=0$$$$\Rightarrow \mathrm{A}=1\mathrm{or}{\mathrm{A}}^2+\mathrm{A}+2=0\left(\mathrm{impossible}\right)$$$$\mathrm{A}=1\Rightarrow {\left(\frac{2}{3}\right)}^{\mathrm{x}}=1\Rightarrow {\mathrm{e}}^{\mathrm{xln}\left(\frac{2}{3}\right)}=1$$$$\Rightarrow \mathrm{xln}\left(\frac{2}{3}\right)=0\Rightarrow \mathrm{x}=0..$$$$$$$${\mathrm{S}}_{\mathbb{R}}=\left\{0\right\}...$$

Commented by Rasheed.Sindhi last updated on 19/Jul/21

$${\left(\frac{8}{18}\right)}^{\mathrm{x}}\stackrel{?}{=}{\left(\frac{2}{3}\right)}^{2\mathrm{x}}$$

Commented by puissant last updated on 19/Jul/21

$${\left(\frac{8}{18}\right)}^{\mathrm{x}}={\left(\frac{4}{9}\right)}^{\mathrm{x}}={\left(\frac{2^2}{3^2}\right)}^{\mathrm{x}}={\left({\left(\frac{2}{3}\right)}^2\right)}^{\mathrm{x}}={\left(\frac{2}{3}\right)}^{2\mathrm{x}}..$$

Commented by Rasheed.Sindhi last updated on 19/Jul/21

$$\mathrm{Ok}\mathrm{sir},\mathrm{thank}\mathrm{you}$$

Commented by otchereabdullai@gmail.com last updated on 19/Jul/21

$$\mathrm{nice}!$$

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