Show that $b = ln(\sqrt[3]{\frac{18}{27-k}})$ for $9 = \frac{k}{3-2e^{-b3}}$ I did:
$$9 = \frac{k}{3-2e^{-b3}} \Leftrightarrow \frac{k}{9} = 3-2e^{b3} \Leftrightarrow \frac{k}{9}-\frac{3}{1} = -2e^{-3b} \Leftrightarrow \frac{\frac{k-27}{9}}{\frac{-2}{1}} \Leftrightarrow -\frac{k-27}{18} = \frac{1}{e^{2b}} \Leftrightarrow -\frac{18}{k-27} = e^{3b} \Leftrightarrow \sqrt[3]{-\frac{18}{k-27}} = e^b \Leftrightarrow b = ln{(\sqrt[3]{-\frac{18}{k-27}})}$$
But my book says that the solution if $b = ln(\sqrt[3]{\frac{18}{27-k}})$
It seems I changed the sign somewhere, but I cannot see where. 
What did I do wrong?

Please note that
$$ -\frac{18}{k-27} = -\frac{18}{-(27-k)} = \not-\frac{18}{\not-(27-k)} = \frac{18}{27-k}$$
Therefore
$$b = \ln\left(\sqrt[3]{-\frac{18}{k-27}}\right) = \ln\left(\sqrt[3]{\frac{18}{27-k}}\right)$$
And hence your solution is the same.
