Cheapest way to find an odd coin among ninety

There are $90$ coins, all of equal weight except one counterfeit that may be either heavier or lighter, you do not know which. Each weighing on a two-pan balance costs $100. What plan identifies the fake and tells whether it is heavy or light at the lowest possible cost in the worst case, and what is that cost?

$500 (five weighings). There are $180$ possibilities and $3^4=81<180\le243=3^5$, so four weighings cannot suffice. Weigh $30$ against $30$: if they balance, the fake is among the other $30$ and the $60$ good coins help four weighings settle it; if not, the $60$ coins each have a known suspected direction, and four weighings resolve these $60\le81$ cases.