A company has determined that demand for its newest netbook model is given by $\ln q−4\ln p+0.002p=7$, where $q$ is the number of netbooks they can sell at a price of p dollars per unit. They have determined that this model is valid for prices $p≥100$. You may find it useful in this problem to know that elasticity of demand is defined to be $E(p)=\frac {dq} {dp}* \frac pq$
how can I find $E(p)$ and where the price gives the maximum revenue?
No, it does not equal that. We have $$ \log(q) = 4\log(p) -0.002p+7.$$ Taking exp of both sides and remembering $e^{a+b}=e^ae^b$ and $e^{a\log(b)}=b^a$ gives $$ q= e^{4\log(p)-0.002p+7} = e^{4\log(p)}e^{-0.002p}e^7 = p^4e^{-0.002p}e^7.$$
You can take the derivative of this using the product rule (although it would be faster to just implicitly differentiate the original equation): $$ \frac{dq}{dp} = 4p^3 e^{-0.002p}e^7 - 0.002e^{-0.002p}p^4e^7.$$ I'll leave it to you to multiply both sides by $p/q$ to get the elasticity (although remember to plug $q$ in as $q(p),$ the function of $p$ we got above, so that the result is a function of price alone).
Revenue is $pq,$ so, using the product rule, $$\frac{d}{dp}(pq) = q+ p\frac{dq}{dp},$$ so setting this equal to zero and solving gives $$ \frac{p}{q}\frac{dq}{dp} = -1.$$ In other words, revenue is maximized when elasticity is $-1.$ So you can plug in your elasticity as a function of price in here and solve for price.