Solving for c

**ctran** · November 24th 2009, 12:31 PM

How would I solve for c for the following equation,

e^(9c/37)= 1/2(e^(36/37)-e^(18/37))

**pickslides** · November 24th 2009, 12:35 PM

$e^{\frac{9c}{37}}= \frac{1}{2}(e^{\frac{36}{37}}-e^{\frac{18}{37}})$

Taking the natural log of both sides you get

$\frac{9c}{37}= \ln\left(\frac{1}{2}(e^{\frac{36}{37}}-e^{\frac{18}{37}})\right)$

Now divide both sides by 9 and multiply both sides by 37.

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