Sunday, June 29, 2008

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:
xlog2 x + log1/4 4 < 4


The solution here is very simple, yet it requires some mathematical knowledge, like what is log1/4 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.

xlog2 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 xlog2 x - 1 < log2 4

Using the log property that says loga xn = nloga 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

1 comment:

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