MathFun4Kids

Solution: Let $a=2^n$, we have $$log_{2^n}(\log_{2^n}(\log_{2^n}2)+\log_{2^n}24-128)=128$$

$$log_{2^n}(\log_{2^n}2)+\log_{2^n}24-128=2^{128n}$$

$$log_{2^n}(\log_{2^n}2^{24})=2^{128n}+128$$

$$log_{2^n}(\dfrac{24}{n})=2^{128n}+128$$

$$\dfrac{1}{n}log_2(\dfrac{24}{n})=2^{128n}+128$$

$$log_2(\dfrac{24}{n})=n(2^{128n}+128)$$

Let $n=3\cdot 2^m$, we have $$3-m=3\cdot 2^m(2^{384\cdot 2^m}+128)$$

Check the range of $m$, we found $-7<m<-5$ and $m=-6$ is a solution. Therefore $n=3\cdot 2^{-6}=\dfrac{3}{64}$, and $a=2^{\frac{3}{64}}$. As $log_ax=a^{256}$, we have

$$log_{2^{\frac{3}{64}}}x=2^{256\cdot{\frac{3}{64}}}=2^{12}$$

$$\dfrac{64}{3}log_{2}x=2^{12}$$

$$log_{2}x=192$$

$$x=2^{192}$$

$$x|1000=2^{192}|1000=4\cdot 2^{190}|1000=4\cdot 1024^{19}|1000=4\cdot 24^{19}|1000$$

$$=96\cdot 24^{18}|1000=96\cdot 13824^6|1000=96\cdot 176^6|1000=96\cdot 30976^3|1000$$

$$=-96\cdot 24^3|1000=-96\cdot 13824|1000=96\cdot 176|1000=16896|1000=\boxed{896}$$

Solution: Assume that the snow was falling at a constant rate of $r$ inches per hour, and the snow plower can remove snow from the streets at a constant rate. The width of the snow plower is $w$ inches. Assume the snow starts to fall $x$ hours before 12 noon. Therefore, we have the following observations:

At 12 o’clock, $x\times r$ inches of snow on the unplowed streets

At 13 o’clock, $(x+1)\times r$ inches of snow on the unplowed streets

At 14 o’clock, $(x+2)\times r$ inches of snow on the unplowed streets

The total volume of snow removed in the first hour would be:

$$6\times w\times\dfrac{x\times r + (x+1)\times r}{2}\ \ \ \ \ {miles\times inch^2}$$

The total volume of snow removed in the second hour would be:

$$3\times w\times\dfrac{(x+1)\times r + (x+2)\times r}{2}\ \ \ \ \ {miles\times inch^2}$$

Because the snow plower removes snow at a constant rate, we have:

$$6\times w\times\dfrac{x\times r + (x+1)\times r}{2}=3\times w\times\dfrac{(x+1)\times r + (x+2)\times r}{2}$$

Solving the above equation, we have $x=\dfrac{1}{2}$. Therefore the snow started to fall at $\boxed{11:30 AM}$.