Soon yi loves to bake, and she is making flaky pastry. soon yi starts with a layer of dough 222 millimeters (\text{mm})(mm)left parenthesis, start text, m, m, end text, right parenthesis thick. the baking process then involves repeatedly rolling out and folding the dough to make layers. each time soon yi rolls and folds the dough, the thickness increases by 8\%8%8, percent. what is the smallest number of times soon yi will have to roll and fold the dough so that the resulting dough is at least 2.5\,\text{mm}2.5mm2, point, 5, start text, m, m, end text thick?

32 times!

Explanation:

Let the original dough thickness be To and the final desired thickness be T in mm.

To=222mm, T = 2.5 m = 2500 mm.

Given that each time Soon Yi rolls and folds the dough, the thickness increases by 8% or 0.08, the final thickness T and the initial thickness To culd be related by the formula:

T = To(1+0.8)^n, where n is the times Soon Yi has to roll the dough.

For instance, after the first roll the thickness will be

T=222(1+0.08)=222*1.08= 239.76 mm

Then the next thickness after the second roll will be

T=239.76*1.08=258.9 mm..., and so on.

By plugging in the given values and evaluating, we have

2500 = 222(1+0.08)^n =222(1.08)^n

2500/222 = 11.2613 = 1.08^n

log(11.2613)=n*log(1.08)

n=log(11.2613)/log(1.08)= 31.46

Thus, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5m thick will be 32.

i think that the correct answer is

c. p equals 5,000 times, 0.3, to the power of the fraction m over 20