I am having trouble with this question. I keep getting X to cancel out somehow. e^(ax) = Ce^(bx), a cannot = b, Find X Please help me!
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Originally Posted by dee18 I am having trouble with this question. I keep getting X to cancel out somehow. e^(ax) = Ce^(bx), a cannot = b, Find X Please help me! Take the natural log of both sides. $\displaystyle \ln(e^{ax}) = \ln(Ce^{bx})$ Using log rules... $\displaystyle ax\ln(e) = bx\ln(Ce)$ More rules... $\displaystyle ax(1) = bx[\ln(C)+\ln(e)]$ Finally $\displaystyle ax=bx(\ln(C)+1)$ Can you solve it now?
$\displaystyle e^{ax} = Ce^{bx}$ $\displaystyle \Rightarrow C = \frac{e^{ax}}{e^{bx}}$ $\displaystyle \Rightarrow C = e^{(a-b)x}$ Now recall that: $\displaystyle \ln{A^x} = x\ln{A}$ Solve for x now.
Sigh, Chop Suey never fails to do the same problem faster and more elegantly...
Thanks guys! I figured it out
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