The half-life of a certain substance is 3.6 days. How long will it take for 20 g of the substance to decay to 7 g?

You know that the half-life of a radioactive nuclide, #t_"1/2"#, tells you the time needed for half of the atoms present in a sample to undergo radioactive decay.

In this case, it takes #3.6# days for any sample of this substance to decay to half of its initial mass.

More specifically, you will have

#"20 g" * 1/2 = "10 g " -># after one half-life, #1 xx "3.6 days"#

#"10 g" * 1/2 = "5 g "-># after two half-lives, #2xx "3.6 days"#

#"5 g" * 1/2 = "2.5 g " -># after three half-lives, #3 xx "3.6 days"#

#vdots#

and so on. In your case, you're interested in finding out how much time must pass for the initial #"20-g"# sample to decay to #"7 g"#. Looking at the numbers, you can say for a fact that it will take more than one half-life and less than two half-lives, so

#overbrace(" 3.6 days ")^(color(purple)("one half-life")) < t < overbrace(" 7.2 days ")^(color(red)("two half-lives"))#

Since the numbers don't allow for a quick calculation, your tool of choice will be the equation

#color(blue)(|bar(ul(color(white)(a/a)A_"t" = A_0 * 1/2^ncolor(white)(a/a)|)))#

Here

#A_"t"# - the mass of the substance that remains undecayed after a period of time #t#
#A_0# - the initial mass of the sample
#n# - the number of half-lives that passed in the period of time #t#

Rearrange the above equation to solve for #n#

#2^n = A_0/A_"t"#

This will be equivalent to

#ln(2^n) = ln(A_0/A_"t")#

#n * ln(2) = ln(A_0/A_"t") implies n = ln(A_0/A_"t")/ln(2)#

Plug in your values to find

#n = ln( (20 color(red)(cancel(color(black)("g"))))/(7color(red)(cancel(color(black)("g"))))) * 1/ln(2) = 1.5146#

Since #n# represents how many half-lives pass in a given period of time, you will have

#t = 1.5146 * "3.6 days" = color(green)(|bar(ul(color(white)(a/a)color(black)("5.5 days")color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one sig figs for the initial mass of the sample and for the mass that remains undecayed.