$$e^{5k}-e^{-5k}=3$$ How do I solve for k using the substitution $y=e^{5k}$ by making it into a quadratic equation?
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For example, what did you get after setting $y=e^{5k}$? – Simply Beautiful Art Oct 16 '17 at 14:38
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Hint: recall that $e^{-5x}=\frac{1}{e^{5x}}\equiv \frac1y$ where the last equivalence is using your substituion.
From here you should know how to go about.
Super Hint: Using the substitution together with the first hint we obtain $$y-\frac1y=3$$ and multiplying with $y$ we obtain $$y^2-1=3y$$ Now you definitely need to solve this one yourself (and keep in mind that $y=0$ cannot be a solution, although in this case it is anyway not a candidate)
b00n heT
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$$e^{5k}-e^{-5k}=3\Longleftrightarrow (e^{5k})^2- 3e^{-5k} -1 = 0\Longleftrightarrow X^2 -3X -1 =0 $$ With $X=e^{5k}$, $$\Delta = 13$$
Then $$X=\frac{3 \pm\sqrt{13}}{2}$$
But $e^{5k} =X>0$ and $\frac{3 -\sqrt{13}}{2}<0$ so $$X=\frac{3 +\sqrt{13}}{2}$$
hence $$k = \frac15 \ln\left( \frac{3 +\sqrt{13}}{2}\right)$$
Guy Fsone
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