News
News Categories

Elon Musk reportedly wants verified users on Twitter to pay for their blue ticks (Updated)

By Kenny Yeo - on 1 Nov 2022, 10:01am

Elon Musk reportedly wants verified users on Twitter to pay for their blue ticks (Updated)

Note: This article was first published on 31 October 2022 and was updated on 1 November 2022 with addition details.

If you haven't heard, Elon Musk now owns Twitter. After much back and forth, Musk finally completed his US$44 billion deal to acquire the social media service.

And Musk wasted no time in enacting his rule on the company. After the deal, he immediately fired all of Twitter's top executives including its CEO, CFO, Head of Legal and Policy, General Counsel and Chief Customer Officer.

And now, Musk is considering a radical move which would force all of its verified users to subscribe to Twitter Blue – Twitter's premium service for US$4.99. Failure to do so will result in losing that blue tick.

Unsurprisingly, the news has not been well received. With one user pointing out that the revenue gains from this move is so inconsequential that it could very well backfire on the service.

Regardless of whether this move goes through, one thing is for sure, the next couple of weeks are going to be very interesting/entertaining for Twitter.

Update on 1 Nov 2022:

A new report says Twitter is planning to increase the fee of Twitter Blue from US$4.99 to US$19.99. Furthermore, it would give users who are currently verified 90 days to subscribe to lose their precious blue ticks.

And that's not all. Employees who are working on this project have been given a deadline of 7 November to roll this feature out or they will be fired.

Even with this new fee, and assuming all of its verified users pay, it would only be enough to cover about 10% of its US$1 billion a year interest.

Source: Platformer, The Verge

Join HWZ's Telegram channel here and catch all the latest tech news!
Our articles may contain affiliate links. If you buy through these links, we may earn a small commission.