### $e$ as a logarithm
-$$\log_e e = 1$$
+$$\operatorname{if} y=e^x, \quad \operatorname{then} x=\log_e y$$
$$\ln x = \log_e x$$
### Differentiating logarithms
-$${d \over dx} \log_b x = {1 \over x \ln b}$$
+$${d(\log_e x)\over dx} = x^-1 = {1 \over x}$$
## Solving $e^x$