How would I solve for c for the following equation,
e^(9c/37)= 1/2(e^(36/37)-e^(18/37))
$\displaystyle e^{\frac{9c}{37}}= \frac{1}{2}(e^{\frac{36}{37}}-e^{\frac{18}{37}})$
Taking the natural log of both sides you get
$\displaystyle \frac{9c}{37}= \ln\left(\frac{1}{2}(e^{\frac{36}{37}}-e^{\frac{18}{37}})\right)$
Now divide both sides by 9 and multiply both sides by 37.