Logarithms and Inequalities

Being independent from Yahoo Answers feels great. I can go to my source, find a great question, and post it here. The topic are somewhat repeating, but there are many more questions. Today's question is about logarithms and inequalities.

Solve:
x^{log₂ x + log_1/4 4} < 4

The solution here is very simple, yet it requires some mathematical knowledge, like what is log_1/4 4. Well, the answer is -1. When the base and the argument are reciprocals, the logarithm's value is -1. Now we can really start solving.

x^{log₂ x - 1} < 4

As you can see, there is a log with base 2 in the exponent. To get rid of the exponent, let's take log base 2 from each side. Since the base is bigger than 1, the inequality sign remains as it is.
log₂ x^{log₂ x - 1} < log₂ 4

Using the log property that says log_a xⁿ = nlog_a x, we can say that:
(log₂ x - 1)log₂ x < 2

Let t = log₂ x

(t - 1)t < 2
t² - t < 2
t² - t - 2 < 0
(t - 2)(t + 1) < 0
-1 < t < 2

Plug the real value of t back:
-1 < log₂ x < 2

By the definition of logs:
1/2 < x < 4

Hope you liked it,
Nadav

nadavs

Limiting Euler

Today's question is about limits and Euler's number, e. This number has many interesting properties, and one of them is the derivative of e^x, which is e^x. Today's question does not deal with the derivative of this function, but with some limits.

Find the limit:
lim_{x -> infinity} (e^x + x)^1/x

This question is definitely not easy, mainly because of that annoying x in there and the 1/x exponent. However, we must never give up. We need to fight this question, so let's fight it with simplicity. Define another variable!

y = (e^x + x)^1/x

Simple, isn't it? All we have to do now is... well... find the limit of y. That didn't help. You can see there is a fraction in the exponent, so if we could just bring it down to a normal fraction, we could use L'Hopital's rule and solve this.

To get the exponent down, we can use a very well known technique: logs. There is a great rule in logarithms which states:
log_a bⁿ = nlog_a b

Using that fantastic rule we can bring the 1/x down and use L'Hopital's rule! To get a nice answer, let's take the logarithm of base e of this function, also written as ln (natural logarithm). So:
ln y = ln (e^x + x)^1/x

Using the logarithm rule mentioned earlier, we can say this equals:
ln (e^x + x) * 1/x
ln (e^x + x) / x

Mr. L'Hopital starts to smile. Now we can find the limit of this function and remember this is the logarithm of the function, so we'll need to consider that.

lim_{x -> infinity} ln y = lim_{x -> infinity} ln (e^x + x) / x

Using L'Hopital's rule (lim_{x -> a} f(x)/g(x) = lim_{x -> a} f'(x)/g'(x)), we can conclude that:
lim_{x -> infinity} ln (e^x + x) / x
is equal to:
lim_{x -> infinity} (e^x + 1) * (1 / (e^x + x)) / 1
lim_{x -> infinity} (e^x + 1) / (e^x + x)

Use L'Hopital's rule twice more:
lim_{x -> infinity} e^x / (e^x + 1)
lim_{x -> infinity} e^x / e^x

Now that looks familiar, doesn't it? This limit equals 1.

Remember: lim_{x -> infinity} ln y = 1, so lim_{x -> infinity} y = e

The limit of the entire function is e. Problem solved.

Hope you enjoyed,
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