Solve, correct to 3 significant figures, the equation e^x + e^2x = e^3x
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$\displaystyle \begin{gathered} e^x \left( {e^{2x} - e^x - 1} \right) = 0 \hfill \\ z \equiv e^x \hfill \\ \Rightarrow z\left( {z^2 - z - 1} \right) = 0 \hfill \\ \end{gathered} $ now solve the quadratic equation and back substitute for x.
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