NO LINKS!!! Please help fill in the blanks. Part 8a

[tex]\\ \tt\hookrightarrow 2(3)^{12t}=1200[/tex]
[tex]\\ \tt\hookrightarrow (3)^{12t}=600[/tex]
[tex]\\ \tt\hookrightarrow log(3)^{12t}=log600[/tex]
[tex]\boxed{\sf log_a b=bloga}[/tex]
[tex]\\ \tt\hookrightarrow 12tlog3=log600[/tex]
[tex]\\ \tt\hookrightarrow t=\dfrac{log600}{12log3}[/tex]
[tex]\\ \tt\hookrightarrow t=\dfrac{2.778}{12(0.477)}[/tex]
[tex]\\ \tt\hookrightarrow t=\dfrac{2.778}{5.724}[/tex]
[tex]\\ \tt\hookrightarrow t=0.485[/tex]
MisterBrainly has already solved part 1, so I'll go over part 2 only.
Check out the screenshot below.
Side note: The portion "Time in years" at the very end could easily be something like "time in days" or "time in "months". It's not clear what time unit your teacher wants. Also, it will depend on how the half-life is set up.