Iodine Solutions

Iodine is a great material for disinfection. It kills bacteria very well, but it's a little burning when put on a wound. Mathematic calculations of iodine solutions are not as burning, but they definitely require some more thought.

A pharmacist has two iodine solutions: one with a concentration of 30% and one with a concentration of 80%. How much of each solution the pharmacist needs to create 8 liters of a 50% solution?

First, we need to understand the situation? Why would a pharmacist need 8 liters of iodine solution? After you answer that in your head, go back to math.

We need two conditions to happen at once: the solution must have 8 liters of solution and 50% iodine, which means 4 liters of iodine. How do we solve that? A system of equations!

Let x by the number of liters of the 80% solution
Let y be the number of liters of the 30% solution

The amount of iodine in each solution is the percent times the volume. In this case, the amount of iodine from the 80% solution is 0.8x and the amount of iodine from the 30% solution is 0.3y. Now let's create those equations:
0.8x + 0.3y = 4
x + y = 8

The 4 comes from the explanation above: 50% of 8 is 4, which means there are 4 liters of iodine in that solution (ouch). Let's solve this system:
0.8x + 0.3y = 4 /*10
x + y = 8

8x + 3y = 40
x + y = 8 /*3

8x + 3y = 40
3x + 3y = 24

Subtract the equations:
5x = 16
x = 16/5 = 3.2 liters

3.2 + y = 8
y = 4.8 liters

Check:
80% * 3.2 = 2.56 liters of iodine
30% * 4.8 = 1.44 liters of iodine
Combined: 4 liters of iodine out of 8 liters of liquid. A 50% solution.

The pharmacist needs 3.2 liters of the 80% solution and 4.8 liters of the 30% solution. That's a lot of iodine.

Have a great weekend,
Nadav

nadavs

Exponential Equations Fury

Yesterday a friend called me and asked me to help her with a system of exponential equations. At first I thought the system is impossible to solve. But then, when I thought all is lost, I started over and used another property of exponentiation. Then it all became clear.

This system is special, so if you know how to solve exponential equations try to solve it first. It gives a great feeling of satisfaction when solved:

2^{x - 1} * 3^{y + 1} = 72 * 6^{y - x + 2}
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}

At first, like most people, you'll be tempted to get everything in terms of powers of 2 and powers of 3 (since 6^x = 2^x * 3^x). However, this will get you nowhere. Instead, notice that both equations have terms with equal exponents, so let's take advantage of that and divide the first equation by the second (I put the factors in an order that will allow me to manipulate them later):

2^{x - 1} / 3^{x - 1} * 3^{y + 1} / 2^{y + 1} = 72 / 3 * 6^{y - x + 2} / 6^{x - y - 2}

Now let's use the exponentation rule that says aⁿ / bⁿ = (a / b)ⁿ. We also need the rule which says aⁿ / a^m = a^{n - m}:
(2/3)^{x - 1} * (3/2)^{y + 1} = 24 * 6^{y - x + 2 - (x - y - 2)}
Using the exponentation rule of 1/aⁿ = a^-n, we can turn the 2/3 into 3/2 and take the opposite of the exponent. Also, let's get the final version of the exponent of 6:
(3/2)^{1 - x} * (3/2)^{y + 1} = 24 * 6^{2y - 2x + 4}

Using the rule that says aⁿ * a^m = a^{n + m}, we can add the exponents of 3/2:
(3/2)^{y - x + 2} = 24 * 6^{2y - 2x + 4}

Look at the exponent of 6, looks familiar, right? Using the rule of (aⁿ)^m = a^nm, we can say:
6^{2y - 2x + 4} = 6^{2(y - x + 2)} = (6²)^{y - x + 2} = 36 ^{y - x + 2}

Now we have a really nice equation with equal exponents on both sides:
(3/2)^{y - x + 2} = 24 * 36^{y - x + 2}

Divide by 36^{y - x + 2}. Using the rules of exponentations mentioned above, this equals:
((3/2) / 36)^{y - x + 2} = 24
(1/24)^{y - x + 2} = 24¹
24^{x - y - 2} = 24¹
x - y - 2 = 1
x - y = 3
x = y + 3

Finally, a connection between x and y. Now let's plug it in one of the original equations to get the real values of x and y:
3^{x - 1} * 2^{y + 1} = 3 * 6^{x - y - 2}
3^{y + 3 - 1} * 2^{y + 1} = 3 * 6^{y + 3 - y - 2}
3^{y + 2} * 2^{y + 1} = 3 * 6¹
3² * 3^y * 2¹ * 2^y = 18
18 * 3^y * 2^y = 18
6^y = 1 = 6⁰
y = 0

Finally, a value. Let's plug that value in the easy equation: x = y + 3. That means x = 3. Problem solved.

Final answer:
x = 3
y = 0

That was long, but it was worth it. Now it's your turn - solve the two towns question using calculus and you will be in the hall of fame. Go ahead, try it.

Nadav

nadavs