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?

Answers

Answer 1

Answer:

Answer:

3 times

Explanation:

When the dough is folded, it increases by a constant factor. We can model the growth of the thickness using the exponential growth model

$T(n)=T_0(1+r)^n$

Where:

Initial thickness, $T_0$ = 2mm

Growth factor, r =8%=0.08

We want to find the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm.

i.e When $T(n)\geq 2.5$ mm$

$2(1+0.08)^n\geq 2.5\n2(1.08)^n\geq 2.5\n$Divide both sides by 2$\n(2(1.08)^n)/(2)\geq (2.5)/(2)\n\n1.08^n\geq 1.25\n\n$Change to logarithm form\nn \geq \log_(1.08)1.25\n\nn\geq (\log 1.25)/(\log 1.08) \n\nn\geq 2.9$

Therefore, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm thick is 3.

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