To solve this, we are going to use the formula: [tex]A=P( \frac{1}{2} ) ^{ \frac{t}{h} } [/tex]
where:
[tex]A [/tex] the final amount of the substance
[tex]P[/tex] is the initial amount of the substance
[tex]t[/tex] is the time
[tex]h[/tex] is the half life
We know from our problem that [tex]A=25[/tex], [tex]P=200[/tex], and [tex]h=3[/tex]. Lets replace those values in our formula:
[tex]A=P( \frac{1}{2} ) ^{ \frac{t}{h} } [/tex]
[tex]25=200( \frac{1}{2})^{ \frac{t}{3} } [/tex]
Notice that [tex]t[/tex] is in the exponent, so we must use logarithms to bring it down:
[tex] \frac{25}{200} =( \frac{1}{2} ) ^{ \frac{t}{3} } [/tex]
[tex]ln( \frac{1}{2} )^{ \frac{t}{3} }=ln( \frac{25}{200} )[/tex]
[tex] \frac{t}{3} ln( \frac{1}{2} )=ln( \frac{25}{250}) [/tex]
[tex] \frac{t}{3}= \frac{ln( \frac{25}{250} )}{ln( \frac{1}{2}) } [/tex]
[tex]t= \frac{3ln( \frac{25}{250}) }{ln( \frac{1}{2}) } [/tex]
[tex]t=9[/tex]
We can conclude that 9 minutes have passed after the substance started to decay for you to have 25 gr of the substance remaining.
