Math Problems, Probability Problems

Problem of the Day: 1/31/13

In the diagram below, the shortest paths from A to B along the gridlines are 6 units long. How many of these paths are there?

20130131-070408.jpg

Solution to yesterday’s problem:

The trick here is that all of the exponents have two as a base. So we have to find a way to take out all the bases and leave only the exponents in our equation. Here’s the equation again:

2^{2n+3}=\frac{2^{n-2}}{2^{2n-2}}

Recall that when you divide two exponents with the same base, the result is the base taken to the difference of the two exponents. For example, \frac{x^5}{x^2}=x^{5-2}=x^3. Here, we can subtract the two exponents as well:

$latex\frac{2^{n-2}}{2^{2n-2}}=2^{(n-2)-(2n-2)}=2^{-n}$

So our new equation is 2^{2n+3}=2^{-n}

In order for the two sides of this equation to be equal, both exponents have to be equal. So here’s where we take out the bases:

2n+3=-n

3n=-3

n=-1

So n = -1.

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