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

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