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

Playing with Logarithms

Logarithms, or logs, are very useful and powerful tools in mathematics. They can be used for many different problems, and all their properties make them very easy to manipulate.

Today I have another question from Yahoo Answers, but I'll take only the good part:

log_a x = c
log_b x = d
Find log_ab x in terms of c and d

This question looks simple. After all, logs have so many properties, there must be one with multiplication of base.

Well, there isn't.

In second thought, there is, but it's not written in any textbook, and to get to it we need to go through many log properties. Shall we start?

First, we can use the log property of reversal of base and argument:
log_a x = 1/log_x a

Now we can write this for both definitions above:
log_x a = 1/log_a x = 1/c
log_x b = 1/log_b x = 1/d

Using the addition property of logs that says: log_x ab = log_x a + log_x b, we can say:
log_x ab = 1 / c + 1 / d = d/cd + c/cd = (c + d) / cd.

Now all we need is to reverse the log back to the one we want:
log_ab x = cd / (c + d)

As you can see, logs don't require too much complicated mathematics, but rather original thinking and creativity.

Do you have any log problems?

Yours,
Nadav

nadavs