chem question please help?

For a first order reaction, the half-life is independent of the initial concentration.

Let n be the number of half-lives taken.
(1/2)ⁿ = 1 - 21%
(0.5)ⁿ = 0.79
log(0.5)ⁿ = log(0.79)
n log(0.5) = log(0.79)
n = log(0.79) / log(0.5)

Half-life = (52.0 min) / [log(0.79) / log(0.5)] = 153 min (to 3 sig. fig.)

====
OR:
Integrated rate equation for first order reaction:
ln([A]/[A]ₒ) = -kt
ln(1 - 21%) = -k (52.0)
Rate constant, k = -ln(0.79) / 52.0 /min

For a first order reaction,
Half-life = ln(2) / k = ln(2) / [-ln(0.79) / 52.0] min = 153 min (to 3 sig. fig.)

There are multiply ways to solve this problem. I prefer to use the integrated rate law for a first order reaction to calculate the rate constant:

ln (A/Ao) = -kt
In this case, the ratio A/Ao = 0.79. So,

ln (0.79) = -k(52.0min)
k = 4.53X10^-3 min^-1

The half-life of the reaction is related to the rate constant as:

t1/2 = ln 2 / k
t1/2 = ln 2 / 4.53X10^-3 min-2 = 153 minutes

you should have this table memorized by now

.. .. .order.. .. .. . non-integrated.. .. .. . ..integrated
.. .. . .. 0.. . .. . ...rate = k x [A]°... .. .. . . ..[At] = -kt + [Ao] 
.. . .. .. 1.. . .. . ...rate = k x [A]¹... .. .. . . ln[At] = -kt + ln[Ao] 
.. . .. . .2... . .. .. .rate = k x [A]²... .. .. . . 1/[At] = +kt + 1/[Ao]

note the difference between non-integrated and integrated
.. non-integrated relates RATE with CONCENTRATION
..... ... integrated relates TIME with CONCENTRATION

*********
you have 1st order, time and concentration, so we use 
.. ln[At] = -kt + ln[Ao]

rearranging
.. (#1).. k = ln([Ao] / [At]) / t.... .. . do you understand how I got this?

and we know that half life occurs when [At] = ½ [Ao] so that
.. .. .. . . t½ = ln([Ao] / ½ [Ao]) / k
.. (#2).. t½ = ln(2) / k

subbing for k.. (#1) into (#2)
.. t½ = ln(2) * t / ln([Ao] / [At])
where
.. [At] = 0.79*[Ao]
.. t = 52.0min

and finally
.. t½ = ln(2) * 52.0min / ln(1/0.79) = -ln(2)*52.0min / ln(0.79) = 153 min 

*********
if you need help with the rearrangements.. you might find these useful
... ln(a) - ln(b) = ln(a/b)
... ln(1/a) = ln(1) - ln(a) = 0 - ln(a) = -ln(a)