Musk Announces Salary Plan of $0 for Twitter Board Members if he Succeeds in Takeover: ’$3M a Year Saved Right There’

It’s not official yet, although Tesla boss Elon Musk wants to buy Twitter outright and take it private, as you have probably already read elsewhere by now. The board of Twitter is attempting to either block this outright or make it much harder or more expensive for Musk to succeed though.

This is likely because many of the left-leaning board members are putting their political bias above the fact that Musk is offering more than the company is currently worth in our view. Additionally, as the broader markets continue to fall, Musk may not have to raise his bid as the valuation of $54.20 per share only starts looking better and better in the face of a potential inflation-caused recession.

Musk has been pushing back against the Twitter board for their attempts to stop him. It is also reported that Twitter board member and co-founder Jack Dorsey is leaving the board, which means collectively the Twitter board doesn’t own a huge chunk of the company.

Musk pointed that out recently as well. Now he is saying that if he is successful he will be cutting the board member’s pay to $0, which he tweeted will save the company money, making his offer even more tempting for non-board member shareholders.

“Board salary will be $0 if my bid succeeds, so that’s ~$3M/year saved right there”

Board salary will be $0 if my bid succeeds, so that’s ~$3M/year saved right there

— Elon Musk (@elonmusk) April 18, 2022

Time will tell how this all shakes out, but to us, it looks like Musk isn’t backing down anytime soon, and Twitter is buying time while they scramble to figure out a way to stop what some see as inevitable.

It’s unclear if cutting board members’ pay will mean they will no longer have their jobs at all or not, or if they’d even want their jobs without pay.

Disclaimer: We are not legal or financial experts and this article is “opinion” and not be taken as trading or financial advice in any way shape or form.