Question 328584
Since 2005 is the base year with 3253 messages, we have

{{{A=3253e^(-.137t)}}}

Let A=100 and solve for t, then add that value to 2005.

{{{100=3253e^(-.137t)}}}

{{{100/3253=e^(-.137t)}}}

{{{ln(100/3253)=-.137t}}}

{{{ln(100/3253)/-.137=25.417}}} years in the future.

2005+25.417=sometime in the year 2030.