To solve, let's apply the formula of half-life which is:

`A = A_o 0.5^(t/t_(1/2))`

where `A` - amount of substance left , `A_o` - original amount of substance,

`t_(1/2) ` - half-life of the substance and `t` - elapsed time

Substitute the given values `A=7g`...

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To solve, let's apply the formula of half-life which is:

`A = A_o 0.5^(t/t_(1/2))`

where `A` - amount of substance left , `A_o` - original amount of substance,

`t_(1/2) ` - half-life of the substance and `t` - elapsed time

Substitute the given values `A=7g` , `A_o = 448g` , and `t_(1/2)=28 days` to the formula.

`7=448*0.5^(t/28)`

Then, isolate t.

To do so, divide both sides by 448.

`7/448=(448*0.5^(t/28))/448`

`7/448=0.5^(t/28)`

Then, take the natural logarithm of both sides of the equation.

`ln(7/448)=ln0.5^(t/28)`

As per property of logarithm, `ln 0.5^(t/28)` becomes `t/28 ln 0.5` .

`ln(7/448)= t/28*ln0.5`

Then, multiply both sides by 28.

`28*ln(7/448)= (t/28*ln0.5)*28`

`28ln(7/448) = t*ln0.5`

Then, divide both sides by ln 0.5.

`(28ln(7/448))/ln0.5= (t*ln0.5)/(ln0.5)`

`(28ln(7/448))/ln0.5=t`

`168=t`

**Hence, it takes 168 days for the 448g Strontium-90 to decay to 7g. **