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Question Number 129951 by Algoritm last updated on 21/Jan/21

Commented by MJS_new last updated on 21/Jan/21

$$2^{x/3}-2^x=2$$$$t=2^{x/3}$$$$t-t^3=2$$$$t\approx -1.52\vee t\approx .761\pm .858\mathrm{i}$$$$$$$$2^{2x/3}+2^{4x/3}+4^x=5$$$$t=2^{x/3}$$$$t^6+t^4+t^2=5$$$$t\approx \pm 1.13\vee t\approx \pm (.648\pm 1.25\mathrm{i})$$$$$$$$2^x-8^x=10(?)$$$$t=2^{x/3}$$$$t^3-t^9=10$$$$t\approx -1.32\vee t\approx -.960\pm .841\vee t\approx -.249\pm 1.25\vee t\approx .661\pm 1.14$$$$$$$$\Rightarrow \mathrm{no}\mathrm{solution}\mathrm{at}\mathrm{all}$$

Answered by SEKRET last updated on 21/Jan/21

$$10$$

Answered by SEKRET last updated on 21/Jan/21

$$2^{\frac{\mathbf{x}}{2}}=\mathbf{a}$$$$\mathbf{a}-{\mathbf{a}}^3=2$$$${\mathbf{a}}^2+{\mathbf{a}}^4+{\mathbf{a}}^6=5$$$$(\mathbf{a}-{\mathbf{a}}^3)({\mathbf{a}}^2+{\mathbf{a}}^4+{\mathbf{a}}^6)=2\cdot 5$$$${\mathbf{a}}^3-{\mathbf{a}}^9=10$$

Commented by MJS_new last updated on 21/Jan/21

$$\mathrm{great}\mathrm{new}\mathrm{method}!$$$$\{\begin{array}{l}{2}^y=9\\ 3^y=4\end{array}$$$$2^y\times 3^y=9\times 4$$$$6^y=36$$$$\Rightarrow y=2$$

Answered by bemath last updated on 21/Jan/21

$$(2^{\frac{\mathrm{x}}{3}}-2^{\mathrm{x}})(2^{\frac{2\mathrm{x}}{3}}+2^{\frac{4\mathrm{x}}{3}}+2^{2\mathrm{x}})=10$$$$2^{\mathrm{x}}+2^{\frac{5\mathrm{x}}{3}}+2^{\frac{7\mathrm{x}}{3}}-2^{\frac{5\mathrm{x}}{3}}-2^{\frac{7\mathrm{x}}{3}}-2^{3\mathrm{x}}=10$$$$2^{\mathrm{x}}-2^{3\mathrm{x}}=10$$$$2^{\mathrm{x}}-8^{\mathrm{x}}=10$$

Commented by MJS_new last updated on 21/Jan/21

$$\mathrm{people},\mathrm{this}\mathrm{is}\mathrm{wrong}!{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}x\mathrm{solving}$$$$\mathrm{both}\mathrm{equations}\mathrm{as}\mathrm{I}\mathrm{showed}.\mathrm{multiplying}$$$$\mathrm{two}\mathrm{equations}\mathrm{is}\mathrm{not}\mathrm{allowed}\mathrm{in}\mathrm{case}\mathrm{the}$$$$\mathrm{variables}\mathrm{are}\mathrm{not}\mathrm{the}\mathrm{same}$$$$\{\begin{array}{l}x=5\\ x=3\end{array}$$$$\stackrel{?}{\Rightarrow }{x}^2=15$$

Commented by liberty last updated on 21/Jan/21

$$\mathrm{do}\mathrm{you}\mathrm{meant}\mathrm{prof}\mathrm{this}\mathrm{question}$$$$\mathrm{is}\mathrm{false}?$$

Commented by MJS_new last updated on 21/Jan/21

$$\mathrm{obviously}$$

Commented by liberty last updated on 21/Jan/21

$$\mathrm{yes}.\mathrm{i}\mathrm{agree}$$

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